PHY 1103

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Contents

1 Vector Algebra 11.1 Definitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.2 Vector Algebra . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1

1.3 Components of Vectors . . . . . . . . . . . . . . . . . . . . . . . . 21.4 Multiplication of Vectors . . . . . . . . . . . . . . . . . . . . . . . 41.5 Vector Field (Physics Point of View) . . . . . . . . . . . . . . . . 61.6 Other Topics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6

2 Electric Force & Electric Field 82.1 Electric Force . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 82.2 The Electric Field . . . . . . . . . . . . . . . . . . . . . . . . . . . 92.3 Continuous Charge Distribution . . . . . . . . . . . . . . . . . . . 122.4 Electric Field Lines . . . . . . . . . . . . . . . . . . . . . . . . . . 18

2.5 Point Charge in E-field . . . . . . . . . . . . . . . . . . . . . . . . 212.6 Dipole in E-field . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22

3 Electric Flux and Gauss Law 253.1 Electric Flux . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 253.2 Gauss Law . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 283.3 E-field Calculation with Gauss Law . . . . . . . . . . . . . . . . . 283.4 Gauss Law and Conductors . . . . . . . . . . . . . . . . . . . . . 31

4 Electric Potential 364.1 Potential Energy and Conservative Forces . . . . . . . . . . . . . 36

4.2 Electric Potential . . . . . . . . . . . . . . . . . . . . . . . . . . . 404.3 Relation Between Electric Field E and Electric Potential V . . . . 454.4 Equipotential Surfaces . . . . . . . . . . . . . . . . . . . . . . . . 48

5 Capacitance and DC Circuits 515.1 Capacitors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 515.2 Calculating Capacitance . . . . . . . . . . . . . . . . . . . . . . . 515.3 Capacitors in Combination . . . . . . . . . . . . . . . . . . . . . . 545.4 Energy Storage in Capacitor . . . . . . . . . . . . . . . . . . . . . 55

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5.5 Dielectric Constant . . . . . . . . . . . . . . . . . . . . . . . . . . 575.6 Capacitor with Dielectric . . . . . . . . . . . . . . . . . . . . . . . 585.7 Gauss Law in Dielectric . . . . . . . . . . . . . . . . . . . . . . . 605.8 Ohms Law and Resistance . . . . . . . . . . . . . . . . . . . . . . 615.9 DC Circuits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 645.10 RC Circuits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 69

6 Magnetic Force 736.1 Magnetic Field . . . . . . . . . . . . . . . . . . . . . . . . . . . . 736.2 Motion of A Point Charge in Magnetic Field . . . . . . . . . . . . 756.3 Hall Effect . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 766.4 Magnetic Force on Currents . . . . . . . . . . . . . . . . . . . . . 78

7 Magnetic Field 817.1 Magnetic Field . . . . . . . . . . . . . . . . . . . . . . . . . . . . 817.2 Parallel Currents . . . . . . . . . . . . . . . . . . . . . . . . . . . 867.3 Amperes Law . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 887.4 Magnetic Dipole . . . . . . . . . . . . . . . . . . . . . . . . . . . . 927.5 Magnetic Dipole in A Constant B-field . . . . . . . . . . . . . . . 937.6 Magnetic Properties of Materials . . . . . . . . . . . . . . . . . . 94

8 Faradays Law of Induction 988.1 Faradays Law . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 988.2 Lenz Law . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 998.3 Motional EMF . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1008.4 Induced Electric Field . . . . . . . . . . . . . . . . . . . . . . . . 104

9 Inductance 1079.1 Inductance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1079.2 LR Circuits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1109.3 Energy Stored in Inductors . . . . . . . . . . . . . . . . . . . . . . 1129.4 LC Circuit (Electromagnetic Oscillator) . . . . . . . . . . . . . . . 1139.5 RLC Circuit (Damped Oscillator) . . . . . . . . . . . . . . . . . . 115

10 AC Circuits 11610.1 Alternating Current (AC) Voltage . . . . . . . . . . . . . . . . . . 11610.2 Phase Relation Between i, V for R,L and C . . . . . . . . . . . . . 11710.3 Single Loop RLC AC Circuit . . . . . . . . . . . . . . . . . . . . . 11910.4 Resonance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12110.5 Power in AC Circuits . . . . . . . . . . . . . . . . . . . . . . . . . 12110.6 The Transformer . . . . . . . . . . . . . . . . . . . . . . . . . . . 123

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11 Displacement Current and Maxwells Equations 12511.1 Displacement Current . . . . . . . . . . . . . . . . . . . . . . . . . 12511.2 Induced Magnetic Field . . . . . . . . . . . . . . . . . . . . . . . . 12711.3 Maxwells Equations . . . . . . . . . . . . . . . . . . . . . . . . . 128

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Chapter 1

Vector Algebra

1.1 Definitions

A vector consists of two components: magnitude and direction .(e.g. force, velocity, pressure)

A scalar consists of magnitude only.(e.g. mass, charge, density)

1.2 Vector Algebra

Figure 1.1: Vector algebra

a + b = b + a

a + (c + d) = (a + c) + d

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1.3. COMPONENTS OF VECTORS 2

1.3 Components of Vectors

Usually vectors are expressed according to coordinate system. Each vector canbe expressed in terms of components.

The most common coordinate system: Cartesian

a = ax + ay + az

Magnitude ofa = |a| = a,

a = a2x + a2y + a2z

Figure 1.2: measured anti-clockwisefrom position x-axis

a = ax + ay

a = a2x + a

2y

ax = acos; ay = a sin

tan =ayax

Unit vectors have magnitude of 1

a =a

|a

|= unit vector along a direction

i j k are unit vectors along x y z directions

a = ax i + ay j + az k

Other coordinate systems:

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1.3. COMPONENTS OF VECTORS 3

1. Polar Coordinate:

Figure 1.3: Polar Coordinates

a = ar r + a

2. Cylindrical Coordinates:

Figure 1.4: Cylindrical Coordinates

a = ar r + a + az z

r originated from nearest point on

z-axis (Point O)

3. Spherical Coordinates:

Figure 1.5: Spherical Coordinates

a = ar r + a + a

r originated from Origin O

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1.4. MULTIPLICATION OF VECTORS 4

1.4 Multiplication of Vectors

1. Scalar multiplication:

If b=ma b,a are vectors; m is a scalarthen b=m a (Relation between magnitude)

bx=m axby=m ay

Components also follow relation

i.e.a = ax i + ay j + az k

ma = max i + may j + maz k

2. Dot Product (Scalar Product):

Figure 1.6: Dot Product

a b = |a| |b| cos

Result is always a scalar. It can be pos-itive or negative depending on .

a b = b aNotice: a b = ab cos = abcosi.e. Doesnt matter how you measureangle between vectors.

i i = |i| |i| cos0

= 1 1 1 = 1i j = |i| |j| cos90 = 1 1 0 = 0

i i = j j = k k = 1i j = j k = k i = 0

If a = ax i + ay j + az kb = bx i + by j + bz k

then a b = axbx + ayby + azbza a = |a| |a| cos0 = a a = a2

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1.4. MULTIPLICATION OF VECTORS 5

3. Cross Product (Vector Product):

If c = a b,then c = |c| = absin

a b = b a !!!

a

b =

b

a

Figure 1.7: Note: How angle is mea-sured

Direction of cross product determined from right hand rule. Also, a b is to a and b, i.e.

a

(a

b) = 0

b (a b) = 0

IMPORTANT:

a a = a asin0 = 0

|i i| = |i| |i| sin0 = 1 1 0 = 0|i j| = |i| |j| sin90 = 1 1 1 = 1

i i = j j = k k = 0i j = k; j k = i; k i = j

a b =

i j kax ay azbx by bz

= (ay bz az by) i+(az bx ax bz) j+(ax by ay bx) k

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1.5. VECTOR FIELD (PHYSICS POINT OF VIEW) 6

4. Vector identities:

a

(b + c) = a

b + a

c

a (b c) = b (c a) = c (a b)a (b c) = (a c)b (a b) c

1.5 Vector Field (Physics Point of View)

A vector field F(x,y,z) is a mathematical function which has a vector outputfor a position input.

(Scalar field

U(x,y,z))

1.6 Other Topics

Tangential Vector

Figure 1.8: dl is a vector that is always tangential to the curve Cwith infinitesimallength dl

Surface Vector

Figure 1.9: da is a vector that is always perpendicular to the surface S withinfinitesimal area da

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1.6. OTHER TOPICS 7

Some uncertainty! (da versus

da)

Two conventions:

Area formed from a closed curve

Figure 1.10: Direction of da determined from right-hand rule

Closed surface enclosing a volume

Figure 1.11: Direction of da going from inside to outside

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Chapter 2

Electric Force & Electric Field

2.1 Electric Force

The electric force between two chargesq1 and q2 can be described byCoulombs Law.

F12 = Force on q1 exerted by q2

F12 =1

40 q1q2

r212 r12

where r12 =r12|r12| is the unit vectorwhich locates particle 1 relative to particle 2.

i.e. r12 = r1

r2

q1, q2 are electrical charges in units of Coulomb(C) Charge is quantized

Recall 1 electron carries 1.602 1019C 0 = Permittivity of free space = 8.85 1012C2/Nm2

COULOMBS LAW:

(1) q1, q2 can be either positive or negative.

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2.2. THE ELECTRIC FIELD 9

(2) If q1, q2 are of same sign, then the force experienced by q1 is in directionaway from q2, that is, repulsive.

(3) Force on q2 exerted by q1:

F21 =1

40 q2q1

r221 r21

BUT:

r12 = r21 = distance between q1, q2

r21 =r21r21

=r2 r1

r21=

r12r12

= r12

F21 = F12 Newtons 3rd Law

SYSTEM WITH MANY CHARGES:

The total force experienced by chargeq1 is the vector sum of the forces on q1

exerted by other charges.

F1 = Force experienced by q1

= F1,2 + F1,3 + F1,4 + + F1,N

PRINCIPLE OF SUPERPOSITION:

F1 =N

j=2F1,j

2.2 The Electric Field

While we need two charges to quantify the electric force, we define the electricfield for any single charge distribution to describe its effect on other charges.

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Total force F = F1 + F2 +

+ FN

The electric field is defined as

limq00

F

q0= E

(a) E-field due to a single charge qi:

From the definitions of Coulombs Law, theforce experienced at location of q0 (point P)

F0,i =1

40 q0qi

r20,i r0,i

where r0,i is the unit vector along the direction from charge qi to q0,

r0,i = Unit vector from charge qi to point P= ri (radical unit vector from qi)

Recall E = limq00

F

q0 E-field due to qi at point P:

Ei =1

40 qi

r2i ri

where ri = Vector pointing from qi to point P,thus ri = Unit vector pointing from qi to point P

Note:

(1) E-field is a vector.

(2) Direction of E-field depends on both position of P and sign of qi.

(b) E-field due to system of charges:

Principle of Superposition:In a system with N charges, the total E-field due to all charges is thevector sum of E-field due to individual charges.

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i.e. E =

i

Ei =1

40

i

qir2i

ri

(c) Electric Dipole

System ofequal and oppositechargesseparated by a distance d.

Figure 2.1: An electric dipole. (Direction ofd from negative to positive charge)

Electric Dipole Moment

p = qd = qdd

p = qd

Example: E due to dipole along x-axis

Consider point P at distance x along the perpendicular axis of the dipole p :

E = E+ + E (E-field (E-field

due to +q) due to q)

Notice: Horizontal E-field components of E+ and E cancel out.

Net E-field points along the axis oppo-site to the dipole moment vector.

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2.3. CONTINUOUS CHARGE DISTRIBUTION 12

Magnitude of E-field = 2E+ cos

E = 2E+ or E magnitude!

1

40 q

r2

cos

But r =

d2

2+ x2

cos =d/2

r

E = 140 p

[x2 + ( d2

)2]3

2

(p = qd)

Special case: When x d

[x2 + (d

2)2]

3

2 = x3[1 + (d

2x)2]

3

2

Binomial Approximation:

(1 + y)n

1 + ny if y

1

E-field of dipole 1

40 p

x3 1

x3

Compare with 1r2

E-field for single charge

Result also valid for point P along any axis with respect to dipole

2.3 Continuous Charge Distribution

E-field at point P due to dq:

d E =1

40 dq

r2 r

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E-field due to charge distribution:

E=

V olume

dE=

V olume

1

40

dq

r2 r

(1) In many cases, we can take advantage of the symmetry of the system tosimplify the integral.

(2) To write down the small charge element dq:

1-D dq = ds = linear charge density ds = small length element2-D dq = dA = surface charge density dA = small area element3-D dq = dV = volume charge density dV = small volume element

Example 1: Uniform line of charge

charge perunit length=

(1) Symmetry considered: The E-field from +z and z directions cancel alongz-direction, Only horizontal E-field components need to be considered.

(2) For each element of length dz, charge dq = dz Horizontal E-field at point P due to element dz =

|d E| cos = 140

dzr2

dEdzcos

E-field due to entire line charge at point P

E =

L/2

L/2

1

40 dz

r2cos

= 2

L/20

40 dz

r2cos

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To calculate this integral:

First, notice that x is fixed, but z, r, all varies.

Change of variable (from z to )

(1)z = x tan dz = x sec2 dx = r cos r2 = x2 sec2

(2) Whenz = 0 , = 0

z = L/2 = 0 where tan 0 =L/2

x

E = 2

40

0

0

x sec2 d

x2 sec2 cos

= 2 40

00

1

x cos d

= 2 40

1x

(sin )0

0

= 2 40

1x

sin 0

= 2

40 1

x L/2

x2 + ( L2

)2

E =1

40 L

x

x2 + ( L2

)2along x-direction

Important limiting cases:

1. x L : E 140

Lx2

But L = Total charge on rod System behave like a point charge

2. L x : E 140

Lx L

2

Ex =

20x

ELECTRIC FIELD DUE TO INFINITELY LONG LINE OF CHARGE

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Example 2: Ring of Charge

E-field at a height z above a ring ofcharge of radius R

(1) Symmetry considered: For every charge element dq considered, there existsdq where the horizontal E field components cancel. Overall E-field lies along z-direction.

(2) For each element of length dz, charge

dq = ds

Linear Circular

charge density length element

dq = R d, where is the anglemeasured on the ring plane

Net E-field along z-axis due to dq:

dE =1

40 dq

r2 cos

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Total E-field =

dE

=2

0140

R dr2 cos (cos = zr )

Note: Here in this case, , R and r are fixed as varies! BUT we want toconvert r, to R, z.

E =1

40 Rz

r3

20

d

E =1

40 (2R)z

(z2 + R2)3/2along z-axis

BUT: (2R) = total charge on the ring

Example 3: E-field from a disk of surface charge density

We find the E-field of a disk byintegrating concentric rings ofcharges.

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Total charge of ring

dq = ( 2r dr Area of the ring

)

Recall from Example 2:

E-field from ring: dE =1

40 dq z

(z2 + r2)3/2

E =1

40

R0

2r dr z(z2 + r2)3/2

=1

40

R0

2zr dr

(z2 + r2)3/2

Change of variable:

u = z2 + r2 (z2 + r2)3/2 = u3/2

du = 2r dr

r dr = 1

2du

Change of integration limit:r = 0 , u = z2

r = R , u = z2 + R2

E =1

40 2z

z2+R2z2

1

2u3/2du

BUT: u3/2du = u1/2

1/2=

2u1/2

E =1

20z (u1/2)

z2+R2z2

=1

20z

1z2 + R2

+1

z

E =

20

1 z

z2 + R2

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2.4. ELECTRIC FIELD LINES 18

VERY IMPORTANT LIMITING CASE:

If R z, that is if we have an infinite sheet of charge with charge den-sity :

E =

20

1 z

z2 + R2

20

1 z

R

E 20

E-field is normal to the charged surfaceFigure 2.2: E-field due to an infi-nite sheet of charge, charge den-sity =

Q: Whats the E-field belows the charged sheet?

2.4 Electric Field Lines

To visualize the electric field, we can use a graphical tool called the electric

field lines.

Conventions:

1. The start on position charges and end on negative charges.

2. Direction of E-field at any point is given by tangent of E-field line.

3. Magnitudeof E-field at any point is proportional to number of E-field linesper unit area perpendicular to the lines.

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2.5. POINT CHARGE IN E-FIELD 21

2.5 Point Charge in E-field

When we place a charge q in an E-fieldE, the force experienced by the charge is

F = qE = ma

Applications: Ink-jet printer, TV cathoderay tube.

Example:

Ink particle has mass m, charge q (q < 0 here)

Assume that mass of inkdrop is small, whats the deflection y of the charge?

Solution:

First, the charge carried by the inkdrop is negtive, i.e. q < 0.

Note: qE points in opposite direction of E.

Horizontal motion: Net force = 0

L = vt (2.1)

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2.6. DIPOLE IN E-FIELD 22

Vertical motion: |qE| |mg|, q is negative,

Net force =

qE = ma (Newtons 2nd Law)

a = qEm

(2.2)

Vertical distance travelled:

y =1

2at2

2.6 Dipole in E-field

Consider the force exerted on the dipole in an external E-field:

Assumption: E-field from dipole doesnt affect the external E-field.

Dipole moment:

p = qd

Force due to the E-field on +veand

ve charge are equal and

opposite in direction. Total ex-ternal force on dipole = 0.

BUT: There is an external torque onthe center of the dipole.

Reminder:

Force F exerts at point P.The force exerts a torque = r F on point P withrespect to point O.

Direction of the torque vector is determined from the right-hand rule.

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Reference: Halliday Vol.1 Chap 9.1 (Pg.175) torqueChap 11.7 (Pg.243) work done

Net torque

direction: clockwisetorque

magnitude:

= +ve + ve

= F d

2 sin + F d

2 sin

= qE d sin = pEsin

= p E

Energy Consideration:

When the dipole p rotates d, the E-field does work.

Work done by external E-field on the dipole:

dW = dNegative sign here because torque by E-field acts to decrease .

BUT: Because E-field is a conservative force field 1 2 , we can define apotential energy (U) for the system, so that

dU = dW For the dipole in external E-field:

dU = dW = pEsin d

U() =

dU =

pEsin d

= pEcos + U01more to come in Chap.4 of notes2ref. Halliday Vol.1 Pg.257, Chap 12.1

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set U( = 90) = 0, 0 = pEcos90 + U0

U0 = 0

Potential energy:

U = pEcos = p E

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Chapter 3

Electric Flux and Gauss Law

3.1 Electric Flux

Latin: flux = to flow

Graphically:Electric flux E represents the number of E-field linescrossing a surface.

Mathematically:

Reminder: Vector of the area A is perpendicular to the area A.

For non-uniform E-field & surface, direction of the area vector A is notuniform.

d A = Area vector forsmall area elementdA

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3.1. ELECTRIC FLUX 26

Electric flux dE = E d AElectric flux of E through surface S: E =

S

E

d A

S

= Surface integral over surface S

= Integration of integral over all area elements on surface S

Example:

E =1

40 2q

r2r =

q20R2

r

For a hemisphere, d A = dA r

E =

S

q20R2

r (dA r) ( r r = 1)

= q20R2

S

dA 2R2

=q0

For a closed surface:

Recall: Direction of area vector d A

goes from inside to outside of closedsurface S.

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3.1. ELECTRIC FLUX 27

Electric flux over closed surface S: E =

S

E d A

S

= Surface integral over closed surface S

Example:

Electric flux of charge q over closedspherical surface of radius R.

E =1

40 q

r2 r =q

40R2 r at the surface

Again, d A = dA r

E =

S

E q

40R2r

d A dA r

=

q

40R2

S dA Total surface area of S = 4R2

E =q

0

IMPORTANT POINT:If we remove the spherical symmetry of closed surface S, the total number of

E-field lines crossing the surface remains the same. The electric flux E

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3.2. GAUSS LAW 28

E =

S

E d A =

S

E d A = q0

3.2 Gauss Law

E =

S

E d A = q0

for any closed surface S

And q is the net electric charge enclosed in closed surface S.

Gauss Law is valid for all charge distributions and all closed surfaces.(Gaussian surfaces)

Coulombs Law can be derived from Gauss Law. For system with high order ofsymmetry, E-field can be easily determined if

we construct Gaussian surfaces with the same symmetry and applies GaussLaw

3.3 E-field Calculation with Gauss Law

(A) Infinite line of charge

Linear charge density: Cylindrical symmetry.E-field directs radially outward from therod.Construct a Gaussian surface S in theshape of a cylinder, making up of acurved surface S1, and the top and

bottom circles S2, S3.

Gauss Law:

S

E d A = Total charge0

=L

0

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3.3. E-FIELD CALCULATION WITH GAUSS LAW 29

S

E d A =

S1

E d A Ed A

+

S2

E d A +

S3

E d A = 0 Ed A

E

S1

dA Total area of surface S1

=L

0

E(2rL) =L

0

E =

20r

(Compare with Chapter 2 note)

E =

20rr

(B) Infinite sheet of charge

Uniform surface charge density:Planar symmetry.E-field directs perpendicular tothe sheet of charge.Construct Gaussian surface S inthe shape of a cylinder (pillbox) of cross-sectional area A.

Gauss Law:

S

E d A = A0

S1

E d A = 0 E d A over whole surface S1

S2

E d A +

S3

E d A = 2EA ( E d A2, E d A3)

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3.3. E-FIELD CALCULATION WITH GAUSS LAW 30

Note: For S2, both E and d A2 point up

For S3, both E and d A3 point down

2EA = A0

E = 20

(Compare with Chapter 2 note)

(C) Uniformly charged sphereTotal charge = QSpherical symmetry.

(a) For r > R:

Consider a spherical Gaussian surface S of

radius r: E d A r

Gauss Law:

S

E d A = Q0

S

E dA = Q0

E

S

dA

surface area of S = 4r2=

Q

0

E =Q

40r2r ; for r > R

(b) For r < R:

Consider a spherical Gaussian surface S ofradius r < R, then total charge included q isproportional to the volume included by S

q

Q=

Volume enclosed by S

Total volume of sphere

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3.4. GAUSS LAW AND CONDUCTORS 31

q

Q=

4/3 r3

4/3 R3 q = r

3

R3Q

Gauss Law:

SE d A = q

0

E

S

dA surface area of S = 4r2

=r3

R31

0 Q

E =1

40 Q

R3r r ; for r R

3.4 Gauss Law and Conductors

For isolated conductors, charges are freeto move until all charges lie outside thesurface of the conductor. Also, the E-field at the surface of a conductor is per-pendicular to its surface. (Why?)

Consider Gaussian surface S of shape of cylinder:

S

E d A = A0

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BUT

S1

E d A = 0 ( E d A )

S3

E d A = 0 ( E = 0 inside conductor )

S2

E d A = E

S2

dA Area of S2

( E d A )

= EA

Gauss Law EA = A0

On conductors surface E = 0

BUT, theres no charge inside conductors.

Inside conductors E = 0 Always!

Notice: Surface charge density on a conductors surface is not uniform.

Example: Conductor with a charge insideNote: This is not an isolated system (because of the charge inside).

Example:

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I. Charge sprayed on a conductor sphere:

First, we know that charges all moveto the surface of conductors.

(i) For r < R:

Consider Gaussian surface S2S2

E d A = 0 ( no charge inside ) E = 0 everywhere.

(ii) For r R:Consider Gaussian surface S1:

S1

E d A = Q0

E

S1

d A 4r2

=Q

0(

For a conductor E d A r

Spherically symmetric

)

E =Q

40r2

II. Conductor sphere with hole inside:

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Consider Gaussian surface S1: Totalcharge included = 0

E-field = 0 inside

The E-field is identical to the case of asolid conductor!!

III. A long hollow cylindrical conductor:

Example:Inside hollow cylinder ( +2q )

Inner radius aOuter radius b

Outside hollow cylinder ( 3q )

Inner radius c

Outer radius d

Question: Find the charge on each surface of the conductor.

For the inside hollow cylinder, charges distribute only on the sur-face. Inner radius a surface, charge = 0and Outer radius b surface, charge = +2q

For the outside hollow cylinder, charges do not distribute only onoutside. Its not an isolated system. (There are charges inside!)

Consider Gaussian surface S inside the conductor:E-field always = 0

Need charge 2q on radius c surface to balance the charge of innercylinder.So charge on radius d surface = q. (Why?)

IV. Large sheets of charge:Total charge Q on sheet of area A,

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Surface charge density =Q

A

By principle of superposition

Region A: E = 0 E = 0

Region B: E =

Q

0A E =

Q

0ARegion C: E = 0 E = 0

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Chapter 4

Electric Potential

4.1 Potential Energy and Conservative Forces

(Read Halliday Vol.1 Chap.12)Electric force is a conservative force

Work done by the electric force F as acharge moves an infinitesimal distance dsalong Path A = dW

Note: ds is in the tangent direction of the curve of Path A.

dW = F ds

Total work done W by force F in moving the particle from Point 1 to Point 2

W =

2

1

F dsPath A

21

= Path Integral

Path A

= Integration over Path A from Point 1 to Point 2.

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4.1. POTENTIAL ENERGY AND CONSERVATIVE FORCES 37

DEFINITION: A force is conservative if the work done on a particle bythe force is independent of the path taken.

For conservative forces,

21

F ds =2

1

F dsPath A Path B

Lets consider a path starting at point1 to 2 through Path A and from 2 to 1through Path C

Work done =

21

F ds +1

2

F dsPath A Path C

= 2

1

F

ds

2

1

F

ds

Path A Path B

DEFINITION: The work done by a conservative force on a particle when itmoves around a closed path returning to its initial position is zero.

MATHEMATICALLY, F = 0 everywhere for conservative force FConclusion: Since the work done by a conservative force F is path-independent,

we can define a quantity, potential energy, that depends only on thepositionof the particle.

Convention: We define potential energy U such that

dU = W =

F ds

For particle moving from 1 to 221

dU = U2 U1 = 2

1

F ds

where U1, U2 are potential energy at position 1, 2.

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Example:

Suppose charge q2moves from point 1to 2.

From definition: U2 U1 = 2

1

F dr

=

r2

r1

F dr ( F

dr )

= r2

r1

1

40

q1q2r2

dr

(

dr

r2= 1

r+ C ) =

1

40

q1q2r

r2

r1

W = U = 140

q1q2

1

r2 1

r1

Note:

(1) This result is generally true for 2-Dimension or 3-D motion.

(2) Ifq2 moves away from q1,then r2 > r1, we have

If q1, q2 are ofsame sign,then U < 0, W > 0(W = Work done by electric repulsive force)

If q1, q2 are ofdifferent sign,then U > 0, W < 0(W = Work done by electric attractive force)

(3) Ifq2 moves towards q1,then r2 < r1, we have

If q1, q2 are ofsame sign,then U 0, W 0

If q1, q2 are ofdifferent sign,then U 0, W 0

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(4) Note: It is the difference in potential energy that is important.

REFERENCE POINT: U(r = ) = 0 U U1 = 1

40q1q2

1

r2 1

r1

U(r) =1

40 q1q2

r

If q1, q2 same sign, then U(r) > 0 for all rIf q1, q2 opposite sign, then U(r) < 0 for all r

(5) Conservation of Mechanical Energy:For a system of charges with no external force,

E = K + U = Constant

(Kinetic Energy) (Potential Energy)

or E = K + U = 0

Potential Energy of A System of Charges

Example: P.E. of 3 charges q1, q2, q3

Start: q1, q2, q3 all at r = , U = 0Step1: Move q1 from to its position U = 0

Step2:

Move q2 from to new position

U =1

40

q1q2r12

Step3:

Move q3 from to new position Total P.E.

U =1

40

q1q2r12

+q1q3r13

+q2q3r23

Step4: What if there are 4 charges?

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4.2. ELECTRIC POTENTIAL 40

4.2 Electric Potential

Consider a charge q at center, we consider its effect on test charge q0

DEFINITION: We define electric potential V so that

V =U

q0=

Wq0

( V is the P.E. per unit charge)

Similarly, we take V(r = ) = 0.

Electric Potential is a scalar.

Unit: V olt(V) = Joules/Coulomb For a single point charge:

V(r) =1

40 q

r

Energy Unit: U = qVelectron V olt(eV) = 1.6 1019

charge of electron

J

Potential For A System of Charges

For a total of N point charges, the po-tential V at any point P can be derivedfrom the principle of superposition.

Recall that potential due to q1 at

point P: V1 = 140 q1r1

Total potential at point P due to N charges:

V = V1 + V2 + + VN (principle of superposition)=

1

40

q1r1

+q2r2

+ + qNrN

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V =1

40

N

i=1qiri

Note: For E, F, we have a sum of vectorsFor V, U, we have a sum of scalars

Example: Potential of an electric dipole

Consider the potential ofpoint P at distance x > d

2

from dipole.

V =1

40

+q

x d2

+q

x + d2

Special Limiting Case: x d1

x d2

=1

x 1

1 d2x

1x

1 d

2x

V =1

40 q

x

1 +

d

2x (1 d

2x)

V =p

40x2(Recall p = qd)

For a point charge E 1r2

V 1r

For a dipole E 1r3

V 1r2

For a quadrupole E 1r4

V 1r3

Electric Potential of Continuous Charge Distribution

For any charge distribution, we write the electri-cal potential dV due to infinitesimal charge dq:

dV =1

40 dq

r

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V =

chargedistribution

1

40 dq

r

Similar to the previous examples on E-field, for the case of uniform chargedistribution:

1-D long rod dq = dx2-D charge sheet dq = dA3-D uniformly charged body dq = dV

Example (1): Uniformly-charged ring

Length of the infinitesimal ring element= ds = Rd

charge dq = ds

= R d

dV = 140

dqr

= 140

R dR2 + z2

The integration is around the entire ring.

V =

ring

dV

=

2

0

1

40 R d

R2 + z2

=R

40

R2 + z2

20

d 2

Total charge on thering = (2R) V =

Q

40

R2 + z2

LIMITING CASE: z R V = Q40

z2

=Q

40|z|

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Example (2): Uniformly-charged disk

Using the principle of superpo-sition, we will find the potentialof a disk of uniform charge den-sity by integrating the potential ofconcentric rings.

dV =1

40

disk

dq

r

Ring of radius x: dq = dA = (2xdx)

V =

R0

1

40 2x dx

x2 + z2

=

40

R

0

d(x2 + z2)

(x2

+ z2

)1/2

V =

20(

z2 + R2

z2)

=

20(

z2 + R2 |z|) Recall:|x| =

+x; x 0x; x < 0

Limiting Case:

(1) If |z| R

z2 + R2 = z21 +

R2

z2 = |z|

1 +

R2

z2

12 ( (1 + x)n 1 + nx if x 1 )

|z|

1 +R2

2z2

(

|z|z2

=1

|z| )

At large z, V 20

R2

2|z| =Q

40|z| (like a point charge)where Q = total charge on disk = R2

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(2) If |z| R

z2 + R2 = R 1 + z2

R2 12

R

1 +z2

2R2

V 20

R |z| + z

2

2R

At z = 0, V =R

20; Lets call this V0

V(z) =

R

20 1 |z

|R +z2

2R2 V(z) = V0

1 |z|

R+

z2

2R2

The keyhere is that it is the difference between potentials of two pointsthat is important. A convenience reference point to compare in this example is thepotential of the charged disk. The important quantity here is

V(z) V0 = |z

|R V0 + &&&&z

2

2R2 V0neglected as z R

V(z) V0 = V0R

|z|

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4.3. RELATION BETWEEN ELECTRIC FIELD E AND ELECTRICPOTENTIAL V 45

4.3 Relation Between Electric Field E and Elec-tric Potential V

(A) To get V from E:Recall our definition of the potential V:

V =U

q0= W12

q0

where U is the change in P.E.; W12 is the work done in bringing chargeq0 from point 1 to 2.

V = V2 V1 = 2

1F

ds

q0

However, the definition of E-field: F = q0 E

V = V2 V1 = 2

1

E ds

Note: The integral on the right hand side of the above can be calculatedalong any path from point 1 to 2. (Path-Independent)

Convention: V = 0 VP = P

E ds

(B) To get E from V:

Again, use the definition of V:

U = q0V = W Work done

However,

W = q0 EElectric force

s

= q0 Es s

where Es is the E-field component alongthe path s.

q0V = q0Ess

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Es = Vs

For infinitesimal s,

Es = dVds

Note: (1) Therefore the E-field component along any direction is the neg-tive derivative of the potential along the same direction.

(2) Ifds E, then V = 0(3) V is biggest/smallest if ds E

Generally, for a potential V(x,y,z), the relation between E(x,y,z) and Vis

Ex = Vx

Ey = Vy

Ez = Vz

x,

y,

zare partial derivatives

For

xV(x,y,z), everything y, z are treated like a constant and we only

take derivative with respect to x.

Example: If V(x,y,z) = x2

y zV

x=

V

y=

V

z=

For other co-ordinate systems

(1) Cylindrical:

V(r,,z)

Er = Vr

E = 1r

V

Ez = Vz

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(2) Spherical:

V(r,,)

Er = Vr

E = 1r

V

E = 1r sin

V

Note: Calculating V involves summation of scalars, which is easier thanadding vectors for calculating E-field.

To find the E-field of a general charge system, we first calculateV, and then derive E from the partial derivative.

Example: Uniformly charged diskFrom potential calculations:

V =

20(

R2 + z2 |z| ) for a point alongthe z-axis

For z > 0, |z| = z

Ez = Vz

=

20 1 z

R2 + z2 (Compare withChap.2 notes)

Example: Uniform electric field(e.g. Uniformly charged +ve and ve plates)

Consider a path going from the veplate to the +ve platePotential at point P, VP can be deducedfrom definition.

i.e. VP V = s

0

E ds (V = Potential ofve plate)=

s0

(E ds) E, ds pointingopposite directions

= E

s0

ds = Es

Convenient reference: V = 0

VP = E

s

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4.4. EQUIPOTENTIAL SURFACES 48

4.4 Equipotential Surfaces

Equipotential surface is a surface on which the potential is constant. (V = 0)

V(r) =1

40 +q

r= const

r = const

Equipotential surfaces arecircles/spherical surfaces

Note: (1) A charge can move freely on an equipotential surface without anywork done.

(2) The electric field lines must be perpendicular to the equipotential

surfaces. (Why?)On an equipotential surface, V = constant V = 0 Edl = 0, where dl is tangentto equipotential surface E must be perpendicular to equipotential surfaces.

Example: Uniformly charged surface (infinite)

Recall V = V0 20

|z|

Potential at z = 0

Equipotential surface means

V = const V0 20

|z| = C |z| = constant

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Example: Isolated spherical charged conductors

Recall:

(1) E-field inside = 0

(2) charge distributed on theoutside of conductors.

(i) Inside conductor:

E = 0 V = 0 everywhere in conductor V = constant everywhere in conductor The entire conductor is at the same potential

(ii) Outside conductor:

V =Q

40r

Spherically symmetric (Just like a point charge.)BUT not true for conductors of arbitrary shape.

Example: Connected conducting spheres

Two conductors con-nected can be seen as asingle conductor

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Potential everywhere is identical.

Potential of radius R1 sphere V1 =

q140R1

Potential of radius R2 sphere V2 =q2

40R2

V1 = V2

q1R1

=q2R2

q1q2

=R1R2

Surface charge density

1 =q1

4R

2

1 Surface area of radius R1 sphere

12

=q1q2

R22

R21=

R2R1

If R1 < R2, then 1 > 2

And the surface electric field E1 > E2

For arbitrary shape conductor:

At every point on the conductor,we fit a circle. The radius of thiscircle is the radius of curvature.

Note: Charge distribution on a conductor does not have to be uniform.

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Chapter 5

Capacitance and DC Circuits

5.1 Capacitors

A capacitor is a system of two conductors that carries equal and oppositecharges. A capacitor stores charge and energy in the form of electro-static field.

We define capacitance as

C =Q

VUnit: Farad(F)

whereQ = Charge on one plate

V = Potential difference between the plates

Note: The C of a capacitor is a constant that depends only on its shape andmaterial.i.e. If we increase V for a capacitor, we can increase Q stored.

5.2 Calculating Capacitance

5.2.1 Parallel-Plate Capacitor

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5.2. CALCULATING CAPACITANCE 52

(1) Recall from Chapter 3 note,

| E| = 0

= Q0A

(2) Recall from Chapter 4 note,

V = V+ V = +

E ds

Again, notice that this integral is independent of the path taken. We can take the path that is parallel to the E-field.

V =

+

E ds

=

+

E ds

=Q

0A

+

ds Length of path taken

=Q

0A

d

(3) C =Q

V=

0A

d

5.2.2 Cylindrical Capacitor

Consider two concentric cylindrical wireof innner and outer radii r1 and r2 re-spectively. The length of the capacitoris L where r1 < r2 L.

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5.2. CALCULATING CAPACITANCE 53

(1) Using Gauss Law, we determine that the E-field between the conductorsis (cf. Chap3 note)

E =1

20

rr =

1

20 Q

Lrr

where is charge per unit length

(2)

V =

+

E ds

Again, we choose the path of integration so that ds r E

V =

r2

r1

E dr = Q20L

r2

r1

drr

ln(r2r1

)

C =Q

V= 20

L

ln(r2/r1)

5.2.3 Spherical Capacitor

For the space between the two conductors,

E =1

40 Q

r2; r1 < r < r2

V =

+

E ds

Choose ds r =r2

r1

1

40 Q

r2dr

=Q

40

1

r1 1

r2

C = 40

r1r2

r2

r1

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5.3. CAPACITORS IN COMBINATION 54

5.3 Capacitors in Combination

(a) Capacitors in Parallel

In this case, its the potential differenceV = Va Vb that is the same across thecapacitor.

BUT: Charge on each capacitor different

Total charge Q = Q1 + Q2

= C1V + C2V

Q = (C1 + C2) Equivalent capacitance

V

For capacitors in parallel: C = C1 + C2

(b) Capacitors in Series

The charge across capacitorsarethe same.

BUT: Potential difference (P.D.) across capacitors different

V1 = Va Vc = QC1

P.D. across C1

V2 = Vc Vb = QC2

P.D. across C2

Potential difference

V = Va Vb= V1 + V2

V = Q (1

C1+

1

C2) =

Q

C

where C is the Equivalent Capacitance

1

C=

1

C1+

1

C2

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5.4. ENERGY STORAGE IN CAPACITOR 55

5.4 Energy Storage in Capacitor

In charging a capacitor, positive chargeis being moved from the negative plateto the positive plate. NEEDS WORK DONE!

Suppose we move charge dq from ve to +ve plate, change in potential energy

dU = V dq = qC

dq

Suppose we keep putting in a total charge Q to the capacitor, the total potentialenergy

U =

dU =

Q0

q

Cdq

U =Q2

2C=

1

2CV2 ( Q=CV)

The energy stored in the capacitor is stored in the electric field between theplates.

Note : In a parallel-plate capacitor, the E-field is constant between the plates.

We can consider the E-field energy

density u =Total energy stored

Total volume with E-field

u =U

Ad

Rectangular volume

Recall

C =0A

d

E =V

d V = Ed

u =1

2(

C0A

d) (

(V)2Ed )2

1

V olume1

Ad

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5.4. ENERGY STORAGE IN CAPACITOR 56

u =1

2

0E2 Energy per unit volume

of the electrostatic fieldcan be generally applied

Example : Changing capacitance

(1) Isolated Capacitor:Charge on the capacitor plates remains constant.

BUT: Cnew =0A

2d=

1

2Cold

Unew =Q2

2Cnew=

Q2

2Cold/2= 2Uold

In pulling the plates apart, work done W > 0

Summary :

Q Q C C/2(V = Q

C) V 2V E E (E = V

d)

12

0E2 = u u U 2U (U = u volume)

(2) Capacitor connected to a battery:Potential difference between capacitor plates remains constant.

Unew =1

2CnewV

2 =1

2 1

2ColdV

2 =1

2Uold

In pulling the plates apart, work done by battery < 0

Summary :

Q

Q/2 C

C/2

V V E E/2u u/4 U U/2

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5.5. DIELECTRIC CONSTANT 57

5.5 Dielectric Constant

We first recall the case for a conductor being placed in an external E-field E0.

In a conductor, charges are free to moveinside so that the internal E-field E setup by these charges

E = E0so that E-field inside conductor = 0.

Generally, for dielectric, the atoms andmolecules behave like a dipole in an E-field.

Or, we can envision this so that in the absence of E-field, the direction of dipole

in the dielectric are randomly distributed.

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5.6. CAPACITOR WITH DIELECTRIC 58

The aligned dipoles will generate an induced E-field E, where |E| < |E0|.We can observe the aligned dipoles in the form of induced surface charge.

Dielectric Constant : When a dielectric is placed in an external E-field E0,the E-field inside a dielectric is induced.E-field in dielectric

E =1

KeE0

Ke = dielectric constant 1

Example :

Vacuum Ke = 1Porcelain Ke = 6.5Water Ke 80Perfect conductor Ke = Air Ke = 1.00059

5.6 Capacitor with Dielectric

Case I :

Again, the charge remains constant after dielectric is inserted.

BUT: Enew =1

KeEold

V = Ed Vnew =1

Ke Vold

C =Q

V Cnew = Ke Cold

For a parallel-plate capacitor with dielectric:

C =Ke0A

d

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5.6. CAPACITOR WITH DIELECTRIC 59

We can also write C =A

din general with

= Ke 0 (called permittivity of dielectric)

(Recall 0 = Permittivity of free space)

Energy stored U =Q2

2C;

Unew =1

KeUold < Uold

Work done in inserting dielectric < 0

Case II : Capacitor connected to a battery

Voltage across capacitor plates remains constant after insertion of dielec-

tric.In both scenarios, the E-field inside capacitor remains constant( E = V /d)

BUT: How can E-field remain constant?ANSWER: By having extra charge on capacitor plates.

Recall: For conductors,

E =

0(Chapter 3 note)

E =Q

0A( = charge per unit area = Q/A)

After insertion of dielectric:

E =E

Ke=

Q

Ke0A

But E-field remains constant!

E = E Q

Ke0A=

Q

0A

Q = KeQ > Q

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5.7. GAUSS LAW IN DIELECTRIC 60

Capacitor C = Q/V C KeCEnergy stored U = 1

2CV2

U

K

eU

(i.e. Unew > Uold)

Work done to insert dielectric > 0

5.7 Gauss Law in Dielectric

The Gauss Law weve learned is applicable in vacuum only. Lets use the capac-itor as an example to examine Gauss Law in dielectric.

Free chargeon plates

Q Q

Induced chargeon dielectric 0 Q

Gauss Law Gauss Law:S

E d A = Q0

S

E d A = Q Q

0

E0 = Q0A

(1) E =Q Q

0A(2)

However, we define E =E0Ke

(3)

From (1), (2), (3) Q

Ke0A

=Q

0A

Q

0A

Induced charge density =Q

A=

1 1

Ke

<

where is free charge density.

Recall Gauss Law in Dielectric:

0

S

E d A = Q Q

E-field in dielectric free charge induced charge

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5.8. OHMS LAW AND RESISTANCE 61

0

S

E d A = Q Q1 1Ke

0

S

E d A = QKe

S

Ke E d A = Q

0

Gauss Lawin dielectric

Note :

(1) This goes back to the Gauss Law in vacuum with E =

E0Ke for dielectric

(2) Only free chargesneed to be considered, even for dielectric where thereare induced charges.

(3) Another way to write: S

E d A = Q

where E is E-field in dielectric, = Ke0 is Permittivity

Energy stored with dielectric:

Total energy stored: U =1

2CV2

With dielectric, recall C =Ke0A

d

V = Ed

Energy stored per unit volume:

ue =U

Ad=

1

2Ke0E

2

and udielectric = Keuvacuum

More energy is stored per unit volume in dielectric than in vacuum.

5.8 Ohms Law and Resistance

ELECTRIC CURRENT is defined as the flow of electric charge through across-sectional area.

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i =dQ

dt

Unit: Ampere (A)= C/second

Convention :

(1) Direction of current is the direction of flow of positive charge.

(2) Current is NOT a vector, but the current density is a vector.

j = charge flow per unit time per unit area

i =

j d A

Drift Velocity :

Consider a current i flowing througha cross-sectional area A:

In time t, total charges passing through segment:

Q = q A(Vdt) Volume of chargepassing through

n

where q is charge of the current carrier, n is density of charge carrierper unit volume

Current: i =Q

t= nqAvd

Current Density: j = nqvd

Note : For metal, the charge carriers are the free electrons inside. j = nevd for metals Inside metals, j and vd are in opposite direction.

We define a general property, conductivity (), of a material as:

j = E

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Note : In general, is NOT a constant number, but rather a function of positionand applied E-field.

A more commonly used property, resistivity (), is defined as =1

E = j

Unit of : Ohm-meter (m)where Ohm () = Volt/Ampere

OHMS LAW:Ohmic materials have resistivity that are independent of the applied electric field.i.e. metals (in not too high E-field)

Example :

Consider a resistor (ohmic material) oflength L and cross-sectional area A.

Electric field inside conductor:

V =

E ds = E L E = VL

Current density: j =i

A

=E

j

=V

L 1

i/A

Vi

= R = LA

where R is the resistance of the conductor.

Note: V = iR is NOT a statement of Ohms Law. Its just a definition forresistance.

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5.9. DC CIRCUITS 64

ENERGY IN CURRENT:

Assuming a charge Q enterswith potential V1 and leaves withpotential V2 :

Potential energy lost in the wire:

U = Q V2 Q V1U = Q(V2

V1)

Rate of energy lost per unit time

U

t=

Q

t(V2 V1)

Joules heating P = i V = Power dissipatedin conductor

For a resistor R, P = i2R =V2

R

5.9 DC CircuitsA battery is a device that supplies electrical energy to maintain a current in acircuit.

In moving from point 1 to 2, elec-tric potential energy increase byU = Q(V2 V1) = Work done by E

DefineE

= Work done/charge = V2

V1

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5.9. DC CIRCUITS 65

Example :

Va = VcVb = Vd assuming

(1) perfect conducting wires.

By Definition: Vc Vd = iRVa Vb = E

E= iR i = ER

Also, we have assumed(2) zero resistance inside battery.

Resistance in combination :

Potential differece (P.D.)

Va Vb = (Va Vc) + (Vc Vb)= iR1 + iR2

Equivalent Resistance

R = R1 + R2 for resistors in series1

R=

1

R1+

1

R2for resistors in parallel

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5.9. DC CIRCUITS 66

Example :

For real battery, there is aninternal resistance thatwe cannot ignore.

E = i(R + r)i =

ER + r

Joules heating in resistor R :

P = i (P.D. across resistor R)= i2R

P =E2R

(R + r)2

Question: What is the value of R to obtain maximum Joules heating?

Answer: We want to find R to maximize P.

dP

dR=

E2(R + r)2

E2 2R

(R + r)3

SettingdP

dR= 0 E

2

(R + r)3[(R + r) 2R] = 0

r

R = 0

R = r

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5.9. DC CIRCUITS 67

ANALYSIS OF COMPLEX CIRCUITS:

KIRCHOFFS LAWS:

(1) First Law (Junction Rule):Total current entering a junction equal to the total current leaving thejunction.

(2) Second Law (Loop Rule):The sum of potential differences around a complete circuit loop is zero.

Convention :

(i)

Va > Vb Potential difference = iR

i.e. Potential drops across resistors

(ii)

Vb > Va Potential difference = +Ei.e. Potential rises across the negative plate of the battery.

Example :

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5.9. DC CIRCUITS 68

By junction rule:i1 = i2 + i3 (5.1)

By loop rule:

Loop A 2E0 i1R i2R + E0 i1R = 0 (5.2)Loop B i3R E0 i3R E0 + i2R = 0 (5.3)Loop C 2E0 i1R i3R E0 i3R i1R = 0 (5.4)

BUT: (5.4) = (5.2) + (5.3)General rule: Need only 3 equations for 3 current

i1 = i2 + i3 (5.1)

3E0 2i1R i2R = 0 (5.2)2E0 + i2R 2i3R = 0 (5.3)

Substitute (5.1) into (5.2) :

3E0 2(i2 + i3)R i2R = 0 3E0 3i2R 2i3R = 0 (5.4)

Subtract (5.3) from (5.4), i.e. (5.4)(5.3)3E0 (2E0) 3i2R i2R = 0

i2 = 54

E0R

Substitute i2 into (5.3) :

2E0 + 5

4 E0

R R 2i3R = 0

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5.10. RC CIRCUITS 69

i3 =

3

8

E0R

Substitute i2, i3 into (5.1) :

i1 =5

4 3

8

E0R

=7

8 E0

R

Note: A negative current means that it is flowing in opposite direction from theone assumed.

5.10 RC Circuits(A) Charging a capacitor with battery:

Using the loop rule:

+E0 iRP.D.across R

QC

P.D.across C

= 0

Note: Direction of i is chosen so that the current represents the rate atwhich the charge on the capacitor is increasing.

E= Ri

dQ

dt+

Q

C1st orderdifferential eqn.

dQEC Q =dt

RC

Integrate both sides and use the initial condition:t = 0, Q on capacitor = 0

Q

0

dQ

EC Q=

t

0

dt

RC

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ln(EC Q)

Q

0=

t

RCt

0

ln(EC Q) + ln(EC) = tRC

ln 1

1 QEC

=

t

RC

11 Q

EC

= et/RC

QEC = 1 et/RC

Q(t) = EC(1 et/RC)

Note: (1) At t = 0 , Q(t = 0) = EC(1 1) = 0

(2) As t , Q(t ) = EC(1 0) = EC= Final charge on capacitor (Q0)

(3) Current:

i =dQ

dt

= EC 1

RC

et/RC

i(t) = ER et/RC i(t = 0) =

ER

= Initial current = i0

i(t ) = 0

(4) At time = 0, the capacitor acts like short circuit when there iszero charge on the capacitor.

(5) As time , the capacitor is fully charged and current = 0, itacts like a open circuit.

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(6) c = RC is called the time constant. Its the time it takes forthe charge to reach (1

1

e) Q

0 0.63Q

0

(B) Discharginga charged capacitor:

Note: Direction of i is chosen so that the current represents the rate atwhich the charge on the capacitor is decreasing.

i = dQdt

Loop Rule:Vc iR = 0

QC

+dQ

dtR = 0

dQ

dt

=

1

RC

Q

Integrate both sides and use the initial condition:t = 0, Q on capacitor = Q0

QQ0

dQ

Q= 1

RC

t0

dt

ln Q ln Q0 = tRC

ln Q

Q0

= t

RC

QQ0

= et/RC

Q(t) = Q0 et/RC

(i = dQdt

) i(t) = Q0RC

et/RC

(At t = 0) i(t = 0) = 1R

Q0C

Initial P.D. across capacitor

i0 =V0R

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At t = RC = Q(t = RC) =1

eQ0 0.37Q0

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Chapter 6

Magnetic Force

6.1 Magnetic Field

For stationary charges, they experienced an electric force in an electric field.For moving charges, they experienced a magnetic force in a magnetic field.

Mathematically, FE = qE (electric force)FB = qv B (magnetic force)

Direction of the magnetic force determined from right hand rule.

Magnetic field B : Unit = Tesla (T)1T = 1C moving at 1m/s experiencing 1N

Common Unit: 1 Gauss (G) = 104T magnetic field on earths surfaceExample: Whats the force on a 0.1C charge moving at velocity v = (10j

20k)ms1 in a magnetic field B = (3i + 4k) 104TF = qv

B

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6.1. MAGNETIC FIELD 74

= +0.1 (10j 20k) (3i + 4k) 104N

= 105

(30 k + 40

i + 60

j + 0)N

Effects of magnetic field is usually quite small.

F = qv B| F| = qvB sin , where is the angle between v and B

Magnetic force is maximum when = 90 (i.e. v B)Magnetic force is minimum (0) when = 0, 180 (i.e. v B)

Graphical representation of B-field: Magnetic field linesCompared with Electric field lines:

Similar characteristics :

(1) Direction of E-field/B-field indicated by tangent of the field lines.

(2) Magnitude of E-field/B-field indicated by density of the field lines.

Differeces :

(1) FE E-field lines; FB B-field lines(2) E-field line begins at positive charge and ends at negative charge; B-

field line forms a closed loop.

Example : Chap35, Pg803 Halliday

Note: Isolated magnetic monopoles do not exist.

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6.2. MOTION OF A POINT CHARGE IN MAGNETIC FIELD 75

6.2 Motion of A Point Charge in Magnetic Field

Since FB v, therefore B-field only changes the directionof the velocity but notits magnitude.

Generally, FB = qv B = q vB , We only need to consider the motioncomponent to B-field.

We have circular motion. Magneticforce provides the centripetal forceon themoving charge particles.

FB = mv2

r

|q| vB = m v2

r

r =mv

|q|Bwhere r is radius of circular motion.

Time for moving around one orbit:

T =2r

v=

2m

qBCyclotron Period

(1) Independent ofv (non-relativistic)

(2) Use it to measure m/q

Generally, charged particles with con-stant velocity moves in helix in the pres-ence of constant B-field.

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6.3. HALL EFFECT 76

Note :

(1) B-field does NO work on particles.(2) B-field does NOT change K.E. of particles.

Particle Motion in Presence of E-field & B-field:

F = qE+ qv B Lorentz Force

Special Case : E B

When |FE| = |

FB|

qE = qvB

v = EB

For charged particles moving at v = E/B , they will pass through thecrossed E and B fields without vertical displacement. velocity selector

Applications :

Cyclotron (Lawrence & Livingston 1934) Measuring e/m for electrons (Thomson 1897) Mass Spectrometer (Aston 1919)

6.3 Hall Effect

Charges travelling in a conducting wire will be pushed to one side of the wire bythe external magnetic field. This separation of charge in the wire is called the

Hall Effect.

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6.3. HALL EFFECT 77

The separation will stop when FB experienced by the current carrier is balancedby the force F

Hcaused by the E-field set up by the separated charges.

Define :

VH = Hall Voltage= Potential difference across the conducting strip

E-field from separated charges: EH =VH

Wwhere W = width of conducting strip

In equilibrium: qEH + qvd B = 0, where vd is drift velocity

VH

W= vdB

Recall from Chapter 5,i = nqAvd

where n is density of charge carrier,A is cross-sectional area = width thickness = W t

VH

W=

i

nqWtB

n = iBqtVH

To determine densityof charge carriers

Suppose we determine n for a particular metal ( q = e), then we can measureB-field strength by measuring the Hall voltage:

B =net

iVH

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6.4. MAGNETIC FORCE ON CURRENTS 78

6.4 Magnetic Force on Currents

Current = many charges moving together

Consider a wire segment, length L,carrying current i in a magnetic field.

Total magnetic force = ( qvd B

force on onecharge carrier) n A L

Total number ofcharge carrierRecall i = nqvdA

Magnetic force on current F = iL B

where L = Vector of which: |L| = length of current segment; direction =direction of current

For an infinitesimal wire segment dl

d F = i dl

B

Example 1: Force on a semicircle current loop

dl = Infinitesimal

arc length element B dl = R d

dF = iRB d

By symmetry argument, we only need to consider vertical forces, dF sin

Net force F =

0

dF sin

= iRB

0

sin d

F = 2iRB (downward)

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Method 2: Write dl in i, j components

dl = dl sin i + dl cos j= R d ( sin i + cos j)

B = B k (into the page) d F = i dl B

= iRB sin d j iRB cos i

F =

0

d F

= iRB

0sin d j +

0

cos d i= 2iRB j

Example 2: Current loop in B-field

For segment2:

F2 = ibB sin(90 + ) = ibB cos (pointing downward)

For segment4:

F2 = ibB sin(90

) = ibB cos (pointing upward)

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For segment1: F1 = iaBFor segment3: F

3= iaB

Net force on the current loop = 0But, net torque on the loop about O

= 1 + 3

= iaB b2

sin + iaB b2

sin

= i abA = area of loop

B sin

Suppose the loop is a coil with N turns of wires:

Total torque = NiAB sin

Define: Unit vector n to represent the area-vector (using right hand rule)

Then we can rewrite the torque equation as

= N iA n BDefine: N iA n = = Magnetic dipole moment of loop

= B

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Chapter 7

Magnetic Field

7.1 Magnetic Field

A moving charge

experiences magnetic force in B-field.

can generate B-field.

Magnetic field B due to moving point charge:

B =0

4 qv r

r2=

0

4 qv r

r3

where 0 = 4 107 Tm/A (N/A2)

Permeability of free space (Magnetic constant)

| B| = 04

qv sin r2

maximum when = 90

minimum when = 0/180

B at P0 = 0 = B at P1B at P2 < B at P3

However, a single moving charge will NOT generate a steady magnetic field.stationary charges generate steady E-field.steady currents generate steady B-field.

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Magnetic field at point P can beobtained by integrating the contribu-tion from individual current segments.(Principle of Superposition)

d B =04

dq v rr2

Notice: dq v = dq

ds

dt= i ds

d B =04

i ds rr2

Biot-Savart Law

For current around a whole circuit:

B =

entirecircuit

d B =

entirecircuit

04

i ds rr2

Biot-Savart Law is to magnetic field asCoulombs Law is to electric field.

Basic element of E-field: Electric charges dqBasic element of B-field: Current element i ds

Example 1 : Magnetic field due to straight current segment

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|ds r| = dzsin = dzsin( ) (Trigonometry Identity)= dz d

r=

d dzd2 + z2

dB =04

i dzr2

dr

=0i

4 d

(d2 + z2)3/2dz

B =

L/2L/2

dB =0id

4

+L/2L/2

dz

(d2 + z2)3/2

B = 0i

4d z

(z2 + d2)1/2

+L/2L/2

B =0i

4d L

( L2

4+ d2)1/2

Limiting Cases : When L d (B-field due to long wire)L2

4+ d2

1/2 L24

1/2=

2

L

B =0i

2d;

direction of B-field determined

from right-hand screw rule

Recall : E =

20dfor an infinite long line of charge.

Example 2 : A circular current loop

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Notice that for every current element ids1, generating a magnetic field d B1at point P, there is an opposite current element ids

2, generating B-field

d B2 so thatd B1 sin = d B2 sin

Only vertical component of B-field needs to be considered at point P.

dB =04

i ds sindsr90

r2

B-field at point P:

B =

aroundcircuit

dB cos consider verticalcomponent

B =

20

0i cos

4r2 ds

Rd

=0i

4 R

r3

20

ds

Integrate around circum-ference of circle = 2R

B =0iR

2

2r3

B =0iR

2

2(R2 + z2)3/2;

direction of B-field determinedfrom right-hand screw rule

Limiting Cases :

(1) B-field at center of loop:

z = 0 B = 0i2R

(2) For z R,

B =0iR

2

2z3

1 + R2

z2

3/2 0iR2

2z3 1

z3

Recall E-field for an electric dipole: E =p

40x3 A circular current loop is also called a magnetic dipole.

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(3) A current arc:

B =

aroundcircuit

dB cos z = 0 = 0 here.

=0i

4 R

r3R = r

when = 0

R = length of arc 0

dsRd

B =

0i

4R

Example 3 : Magnetic field of a solenoidSolenoid is used to produce a strong and uniform magnetic field inside itscoils.

Consider a solenoid of length L consisting of N turns of wire.

Define: n = Number of turns per unit length =N

L

Consider B-field at distance d from thecenter of the solenoid:For a segment of length dz, number ofcurrent turns = ndz

Total current = nidz

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7.2. PARALLEL CURRENTS 86

Using the result from one coil in Example 2, we get B-field from coils oflength dz at distance z from center:

dB =0(nidz)R

2

2r3

However r =

R2 + (z d)2

B =

+L/2

L/2

dB(Integrating over theentire solenoid)

=0niR

2

2

+L/2L/2

dz

[R2 + (z d)2]3/2

B =0ni

2

L2 + d

R2 + ( L2

+ d)2+

L2

dR2 + ( L

2 d)2

along negative z direction

Ideal Solenoid :

L Rthen B =

0ni

2[1 + 1]

B = 0ni ;direction of B-field determinedfrom right-hand screw rule

Question : What is the B-field at the end of an ideal solenoid? B=0ni2

7.2 Parallel Currents

Magnetic field at point P B due to twocurrents i1 and i2 is the vector sum ofthe B fields B1, B2 due to individual cur-rents. (Principle of Superposition)

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7.2. PARALLEL CURRENTS 87

Force Between Parallel Currents :

Consider a segment of length L on i2 :

B1 =0i12d (pointing down) B2 =

0i12d (pointing up)

Force on i2 coming from i1:

| F21| = i2L B1 = 0Li1i22d

= | F12| (Defn of ampere, A) Parallel currents attract, anti-parallel currents repel.

Example : Sheet of current

Consider an infinitesimal wire of width dx at position x, there exists anotherelement at x so that vertical B-field components of B+x and Bx cancel. Magnetic field due to dx wire:

dB =0 di

2rwhere di = i

dxa

Total B-field (pointing along x axis) at point P:

B =

+a/2

a/2

dB cos =

+a/2

a/2

0i

2a dx

r cos

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7.3. AMPERES LAW 88

Variable transformation (Goal: change r, x to d, , then integrate over ):

d = r cos r = d sec x = d tan dx = d sec2 d

Limits of integration: 0 to 0, where tan 0 = a2d

B =0i

2a

00

d sec2 d

d sec cos

=0i

2a

00

d

B = 0i0a

= 0i

atan1

a2d

Limiting Cases :

(1) d atan =

a

2d a

2d

B =0i

2aB-field due toinfinite long wire

(2) d a (Infinite sheet of current)tan =

a

2d =

2

B =0i

2aConstant!

Question : Large sheet of opposite flowing currents.

Whats the B-field between & outside the sheets?

7.3 Amperes Law

In our study of electricity, we notice that the inverse square force law leadsto Gauss Law, which is useful for finding E-field for systems with high level ofsymmetry.For magnetism, Gauss Law is simple

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7.3. AMPERES LAW 89

S

S

B

d A = 0

There is no mag-netic monopole.

A more useful law for calculating B-field for highly symmetric situations is theAmperes Law:

C

C

B ds = 0i

C

= Line intefral evaluated around a closed loop C (Amperian curve)

i = Net current that penetrates the area bounded by curve C (topological property)

Convention : Use the right-hand screw rule to determine the signof current.

C

B ds = 0(i1 i3 + i4 i4)

= 0(i1 i3)

Applications of the Amperes Law :

(1) Long-straight wire

Construct an Amperiancurve of radius d:

By symmetry argument, we know B-field only has tangential compo-nent

C

B ds = 0i

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7.3. AMPERES LAW 90

Take ds to be the tangential vector around the circular path:

B ds = B dsB

C

ds Circumferenceof circle = 2d

= 0i

B(2d) = 0i

B-field due to long,straight current

B =0i

2d(Compare with 7.1 Example 1)

(2) Inside a current-carrying wire

Again, symmetry argumentimplies that B is tangentialto the Amperian curve andB B(r)

Consider an Amperian curve of radius r(< R)

C

B ds = B ds = B(2r) = 0iincluded

But iincluded cross-sectional area of C

iincluded

i=

r2

R2

iincluded =r2

R2i

B = 0i2R2

r r

Recall: Uniformly charged infinite long rod

(3) Solenoid (Ideal)

Consider the rectangularAmperian curve 1234.

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7.3. AMPERES LAW 91

C

B ds =

1

B ds +

&&&

&&2

B ds +

&&&

&&3

B ds +

&&&

&&4

B ds

2

=

4

= 0

B ds = 0 inside solenoid

B = 0 outside solenoid3

= 0 B = 0 outside solenoid

C

B ds =

1

B ds = Bl = 0itotBut itot = nlNumber of coils included

i

B = 0ni

Note :

(i) The assumption that B = 0 outside the ideal solenoid is onlyapproximate. (Halliday, Pg.763)

(ii) B-field everywhere inside the solenoid is a constant. (for idealsolenoid)

(4) Toroid (A circular solenoid)

By symmetry argument, the B-field lines form concentric circles insidethe toroid.Take Amperian curve C to be a circle of radius r inside the toroid.

C

B ds = B

C

ds = B 2r = 0(N i)

B =0Ni

2rinside toroid

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7.4. MAGNETIC DIPOLE 92

Note :

(i) B = constant inside toroid(ii) Outside toroid:

Take Amperian curve to be circle of radius r > R.

C

B ds = B

C

ds = B 2r = 0 iincl = 0

B = 0

Similarly, in the central cavity B = 0

7.4 Magnetic Dipole

Recall from 6.4, we define the magnetic dipole moment of a rectangularcurrent loop

= NiAn

where n = area unit vector with directiondetermined by the right-hand rule

N = Number of turns in current loopA = Area of current loop

This is actually a general definition of a magnetic dipole, i.e. we use it forcurrent loops of all shapes.

A common and symmetric example: circular current.

Recall from 7.1 Example 2, magneticfield at point P (height z above the ring)

B =0iR

2n

2(R2 + z2)3/2=

0

2(R2 + z2)3/2

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7.5. MAGNETIC DIPOLE IN A CONSTANT B-FIELD 93

At distance z R,

B = 0

2z3E = p

40z3

due to magnetic dipole due to electric dipole(for z R) (for z d)

Also, notice = magnetic dipole moment

Unit: Am2

J/T

0 = Permeability of free space

= 4 107Tm/A

7.5 Magnetic Dipole in A Constant B-field

In the presence of a constant magnetic field, we have shown for a rectangular

current loop, it experiences a torque = B . It applies to any magneticdipole in general.

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7.6. MAGNETIC PROPERTIES OF MATERIALS 94

External magnetic field aligns the magneticdipoles.

Similar to electric dipole in a E-field, we can con-sider the work done in rotating the magnetic di-pole. (refer to Chapter 2)

dW = dU, where U is potential energy of dipole

U = B

Note :

(1) We cannot define the potential energy of a magnetic field in general.However, we can define the potential energy of a magnetic dipole in a

constant magnetic field.(2) In a non-uniform external B-field, the magnetic dipole will experience

a net force (not only net torque)

7.6 Magnetic Properties of Materials

Recall intrinsic electric dipole in molecules:

Intrinsic dipole (magnetic) in atoms:

In our classical model of atoms, electronsrevolve around a positive nuclear.

Current i =e

P

, where P is period of one orbit around nucleus

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P =2r

v, where v is velocity of electron

Orbit magnetic dipole of atom:

= iA = ev

2r

(r2) =

erv

2

Recall: angular momentum of rotation l = mrv

=e

2m l

In quantum mechanics, we know that

l is quantized, i.e. l = N

h

2

where N = Any positive integer (1,2,3, ... )

h = Plancks constant (6.63 1034J s) Orbital magnetic dipole moment

l =eh

4m Bohrs magneton B=9.271024J/T

N

There is another source of intrinsic magnetic dipole moment inside an atom:

Spin dipole moment: coming from the intrinsic spin of electrons.

Quantum mechanics suggests that e are alwaysspinning and its either an upspin or a down spin

e = 9.65 1027 BSo can there be induced magnetic dipole?

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Recall: for electric field

Edielectric = KeEvacuum ; Ke 1For magnetic field in a material:

Bnet = B0 + BM

appliedB-field

B-field producedby induced dipoles

In many materials (except ferromagnets),

BM

B0

Define :BM = m B0

m is a number called magnetic susceptibility.

Bnet = B0 + m B0

= (1 + m) B0

Bnet = m B0 ; m = 1 + m

Define : m is a number called relative permeability.

One more term ......

Define : the Magnetization of a material:

M =d

dVwhere is magnetic dipolemoment, V is volume

(or, the net magnetic dipole moment per unit volume)

In most materials (except ferromagnets),

BM = 0 M

Three types of magnetic materials:

(1) Paramagnetic:

m 1(m 0)

,induced magnetic dipoles aligned

with the applied B-field.

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e.g. Al (m.

= 2.2 105), Mg (1.2 105), O2 (2.0 106)

(2) Diamagnetic:

m 1(m 0) ,

induced magnetic dipoles alignedopposite with the applied B-field.

e.g. Cu (m 1 105), Ag (2.6 105), N2 (5 109)(3) Ferromagnetic:

e.g. Fe, Co, NiMagnetization not linearly proportionalto applied field.

BnetBapp

not a constant (can be as

big as 5000 100, 000)

Interesting Case : Superconductors

m = 1

A perfect diamagnetic.NO magnetic field inside.

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Chapter 8

Faradays Law of Induction

8.1 Faradays Law

In the previous chapter, we have shown that steady electric current can givesteady magnetic field because of the symmetry between electricity & magnetism.

We can ask: Steady magnetic field can give steady electric current. OR Changing magnetic field can give steady electric current.

Define :

(1) Magnetic flux through surface S:

m =

S

B d A

Unit of m : Weber (Wb)1Wb = 1Tm2

(2) Graphical:

m = Number of magnetic field lines passing through surface S

Faradays law of induction:

Induced emf |E| = Ndmdt

where N = Number of coils in the circuit.

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8.2. LENZ LAW 99

B = Constant B = Constant B = Constant B = ConstantA = Constant A = Constant dB/dt = 0 A = Constant

dA/dt = 0 A = Constant dA/dt = 0E= 0 |E| > 0 |E| > 0 |E| > 0

Note : The induced emf drives a current throughout the circuit, similar to thefunction of a battery. However, the difference here is that the induced emfis distributed throughout the circuit. The consequence is that we cannotdefine a potential difference between any two points in the circuit.

Suppose there is an induced current in the loop, can wedefine VAB?

Recall:

VAB = VA

VB = iR > 0

VA > VBGoing anti-clockwise (same as i),

If we start from A, going to B, then we get VA > VB.If we start from B, going to A, then we get VB > VA.

We cannot define VAB !!

This situation is like when we study the interior of a battery.

A battery

The loop

provides the energy needed to drive thecharge carriers around the circuit by

chemical reactions.

changing magnetic flux.

sources of emf non-electric means

8.2 Lenz Law

(1) The flux of the magnetic field due to induced current opposes the changein flux that causes the induced current.

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8.3. MOTIONAL EMF 100

(2) The induced current is in such a direction as to oppose the changes thatproduces it.

(3) Incorporating Lentz Law into Faradays Law:

E= N dmdt

Ifdm

dt> 0, m Eappears Induced currentappears.

B-field due toinduced current

change in m so that= m

(4) Lenz Law is a consequence from the principle of conservation of energy.

8.3 Motional EMF

Lets try to look at a special case when the changing magnetic flux is carried bymotion in the circuit wires.

Consider a conductor of length L movingwith a velocity v in a magnetic field B.

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Hall Effect for the charge carriers in the rod:

FE + FB = 0

qE+ qv B = 0 (where E is Hall electric field) E = v B

Hall Voltage inside rod:

V = L

0

E dsV = EL

Hall Voltage : V = vBL

Now, suppose the moving wire slides withoutfriction on a stationary U-shape conductor.The motional emf can drive an electric cur-rent i in the U-shape conductor. Power is dissipated in the circuit. Pout = V i (Joules heating)(see Lecture Notes Chapter 4)

What is the source of this power?Look at the forces acting on the conducting rod:

Magnetic force:Fm = iL BFm = iLB (pointing left)

For the rod to continue to move at constant velocity v, we need to applyan external force:

Fext = Fm = iLB (pointing right)

Power required to keep the rod moving:

Pin = Fext v= iBLv

= iBLdx

dt

= iBd(xL)

dt( xL = A, areaenclosed by circuit)

= id(BA)

dt

( BA = m, magnetic flux)

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Since energy is not being stored in the system,

Pin + Pout = 0

iV + idm

dt= 0

We prove Faradays Law V = dmdt

Applications :

(1) Eddy current: moving conductors in presence of magnetic field

Induced current

Power lost in Joules heatingE2

R

Extra power input to keep moving

To reduce Eddy currents:

(2) Generators and Motors:Assume that the circuit loop is rotating at a constant angular velocity, (Source of rotation, e.g. steam produced by burner, water fallingfrom a dam)

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Magnetic flux through the loop

Number of coils

B = N

loop

B d A = N BA cos

changes with time! = t

B = N BA cos t

Induced emf: E= dBdt

= N BA ddt

(cos t)

= NBA sin t

Induced current: i =ER

=NBA

Rsin t

Alternating current (AC) voltage generator

Power has to be provided by the source of rotation to overcome thetorque acting on a current loop in a magnetic field.

=

Ni A B

= NiAB sin

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8.4. INDUCED ELECTRIC FIELD 104

The net effect of the torque is to oppose the rotation of the coil.

An electric motor is simply a generatoroperating in reverse. Replace the load resistance R witha battery of emfE.

With the battery, there is a current in the coil, and it experiences atorque in the B-field. Rotation of the coil leads to an induced emf, Eind, inthe direction opposite of that of the battery. (Lenz Law)

i =E Eind

R

As motor speeds up, Eind , i mechanical power delivered = torque delivered = NiAB sin In conclusion, we can show that

Pelectric = i2R + Pmechanical

Electric power input Mechanical power delivered

8.4 Induced Electric Field

So far we have discussed that a change in mag-netic flux will lead in an induced emf distributedin the loop, resulting from an induced E-field.

However, even in the absence of the loop (so that there is no induced current),the induced E-field will still accompany a change in magnetic flux.

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Consider a circular path in a regionwith changing magnetic field.

The induced E-field only has tangential components. (i.e. radial E-field = 0)Why?

Imagine a point charge q0 travelling around the circular path.Work done by induced E-field = q0Eind force

2rdistanceRecall work done also equals to q0E, where Eis induced emf

E= Eind2r

Generally,

E=

Eind ds

where

is line integral around a closed loop, Eind is induced E-field, s istangential vector of path.

Faradays Law becomes

C

Eind ds = ddt

S

B d A

L.H.S. = Integral around a closed loop CR.H.S. = Integral over a surface bounded by C

Direction ofd A determined by direction of line integration C (Right-Hand Rule)

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Regular E-field Induced E-field

created by charges created by changing B-field

E-field lines start from +ve and endon ve charge E-field lines form closed loops

can define electric potential so thatwe can discuss potential difference

between two points

Electric potential cannot be defined(or, potential has no meaning)

Conservative force field Non-conservative force field

The classification of electric and magnetic effects depend on the frame of referenceof the observer. e.g. For motional emf, observer in the reference frame of themoving loop, will NOT see an induced E-field, just a regular E-field.(Read: Halliday Chap.33-6, 34-7)

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Chapter 9

Inductance

9.1 Inductance

An inductor stores

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PHY 1103