Chapter V Magnetostatics

Recommended problems: 5.1, 5.4, 5.8, 5.9, 5.10, 5.11, 5.13, 5.14, 5.15, 5.22,

5.23, 5.24, 5.27, 5.44, 5.45, 5.46.

The Lorentz Force Law

It is known that any stationary charge produces an e.f.

If a charge q is placed in an e.f. It will experiences an e. force according to

)1(EqF

Now a moving charge produces a m.f, B and any moving charge

in a magnetic field experiences a m. force according to.

LawLorentsBvqF )2(

From Eq.(2), it is clear that the m.force is perpendicular to both v & B.

Now since dt

sdv

0sdFm

The magnetic force doesn't do any work in a moving charge.

Motion of a Charged Particle in M. Field

It is shown that the magnetic force doesn’t do any work

If the magnetic force is the only force acting on a particle and knowing that

KWnet 0K

The speed is constant the acceleration is due to the change in direction

only the force is centripetal, i.e., the motion of the charge is circular.

Example 5.1 Describet the motion of a charged particle in a uniform m.field.

kBBLet ˆ

With B is constant.

2

2

dt

rdmBvqFNow m

jxBiyB

B

zyx

kji

BvBut ˆˆ

00

ˆˆˆ

kzjyixmjxBiyBq ˆˆˆˆˆ

Solution.

)3(yqBxm

)4(xqBym

)5(0zm

Integrating Eqs. 3-4 with respect to time, we get, respectively

)6(1CqByxm

)7(2CqBxym

)8(3Cz

Substituting for from Eq. (7) into Eq. (3) y

Cx

m

qBx

2

)9(22 axx m

qBwith

Solving Eq.(9) )10(cos otRax

Differentiating Eq. (10) w.r.t time and substituting into Eq. (6)

CytR o sin

)11(sin otRby

Squaring Eqs(10+11) and then adding we get

)12(222Rbyax

Then the path of the motion is a circle of radius R and centered at (a,b).

Now for the motion is helix with its axis in the direction of B. constantz

Now from Eqs.(10&11) we have

otRx sin otRy cos

Squaring the above two equations and then adding we get

22222 vwRyx qB

mvvR

Example 5.2 A particle of mass m and charge q starts from rest at the origin in

the presence of both E and B such that and . Describe the

motion of the particle.

kEE ˆ

iBB ˆ

Solution.

kyBjzB

B

zyx

kji

Bv ˆˆ

00

ˆˆˆ

BvqEqFFF me

)13(ˆˆˆ kyjzqBkqErm

)14(zqBym

)15(yqBqEzm

)16(0x

From Eq.(16), and since the particle stars from rest, then 0x

Integrating Eq.(14), and since the particle stars from rest, we get

)17(1 zCzy

Substituting Eq.(17) back into Eq.(15) we get

zm

qEz 2

m

qEzz 2

)18(cos otARz

Integrating Eq.(15), and since the particle stars from rest, we get

)19(2 ytm

qECyt

m

qEz

Differentiating Eq.(18) we get )20(sin otAz

Equating Eqs.(19 & 20) we obtain

ytm

qEtA o sin

)21(sin tm

qEtAy o

Squaring Eqs. (18 & 21) and then adding we get

)22(222RRztRy

Which is the equation of a circle of radius R and center positioned at (0,Rt, R)

travels along the y-axis at constant speed. The curve generated in this way is

called cycloid with.

)23(B

ER

)24(

B

ERv

Current

A moving charge in one direction produces current I with

)25(dvAnedt

dqI

With A is the cross sectional area of the wire, n is the No. of free charges per

unit volume, and vd is the drift velocity.

The magnetic force on current-carrying wire is

)26( BldIIdtBvdqBvFm

Example 5.3 A rectangular loop of wire, supporting

a mass m, hangs vertically with one end in a

uniform m.f. B. Find I such that m be in equilibrium.

Solution. For the mass to be in equilibrium we have

0gm FF

jmgFF gmˆ

jIaBkiIaBBlIFBut mˆ)ˆˆ(

IaBmgaB

mgI

The Current Density

The current density J is defined such that

)27( AdJI

Or from Eq.(25) we have

)28(dvJ

Now the magnetic force can be written as

)29( dBJdBvdqBvFm

Let us calculate the current enering a closed surface S. We have from Eq.(27)

S

dSnJI ˆ

The minus sign because n is outward.

Using the divergence theorem, we get

)30(ˆ dJdSnJIS

But the current is equal to the rate at which charge is transported into the

volume, i.e.,

)31(V

ddt

d

dt

dqI

Since the volume is constant and knowing that is a function of position as well

as of time, we get

)32(

Vd

tI

Equating Eqs. (30 & 32) and since the volume is arbitrary we get

)33(t

J

Continuity Equation

An outflow of current through a volume decreases the charge inside that

volume

The Biot-Savart Law

The m.f at a point due to a wire of steady current I is given by

)34(4 3

ldIB o

Is an element along the wire. ld

is a vector from the source to the point, and rr

27 /104 ANo is the magnetic permeability of free space.

dl

I

P

Example 5.5 Find the m.f a distance s

from a long straight wire carrying a

steady current I.

dz

z

I

P

s

Solution. We begin by identifying an

element of the wire, having length

dl=dz. Now

kdzld ˆ

ssr ˆ

kzr ˆ

kzssrr ˆˆ

33

ˆˆˆ

44

kzsskdzIldIB oo

2

12

3224

l

l

o

zs

dzIsB

22222 2

3Using

xaa

x

xa

dx

2

322

ˆ

4zs

dzIsB o

221

1

222

2

22 44

2

1

23

sl

l

sl

l

s

I

zs

z

s

IB o

l

l

o

If the wire in infinite we have 21 ll

s

I

sls

IlB oo

l

22lim

222

To find the force between two parallel wires we have for

the force on wire 1 due to wire 2

F1

a

I2 I1

B2

21112 BlIF

kd

IBBut o ˆ

2

22

is the m.f due to I2 at the position of I1

jld

IIF o ˆ

2

2112

The force is attractive if the currents are in the same direction

and repulsive if the currents are in opposite directions.

Example 5.6 Find the m.f due a distance z from the

center of a circular loop of radius R which carries a

steady current I.

Solution. We begin by identifying an element of the

wire, having length dl’ with

jiRdRdld ˆcosˆsinˆ

kzr ˆ jiRsRr ˆsinˆcosˆ

jRiRkzrr ˆsinˆcosˆ

zRR

dRdR

kji

ld

sincos

0cossin

ˆˆˆ

dkRjRziRz ˆˆsinˆcos 2

2

0 22

2

32

3

ˆˆsinˆcos

44zR

dkRjRziRzIldIB oo

dl’ I

P

z

R

0 yx BB 2

322

2

2 zR

IRB o

z

Note that at the origin of the loop (z=0 ) we get R

IB o

z2

Divergence and Curl of B

From Biot-Savart law we have

34

ldIB o

34

ldIB o

ABBABABut

ld

IldIB oo

33 44

0&1

3

But

ld

Ild

IB oo

1

4

1

4

)35(0 B

Again using Biot-Savart law we have

34

ldIB o

AdJI

&

)36(

4 3

d

rJB o

drJB o

34

BACCABCBAgU

sin

JJrJ

333

0&4Using3

J

drrrJB o

)37(rJB o

Ampere’s Law

Using Stoke's theorem we have

ldBSdB

rJB o

From Eq.(37) we have SdJldB o

)38(enco IldB

Where Ienc is the total current crossing the surface enclosed by the closed loop.

Like Gauss’s law, Ampere’e law is always true for steady current, but it is not

always useful. Only when the symmetry of the problem enables you to pull B

out of the integral , can you calculate B using Ampere’s law. ldB

The current configurations that can be handled by Ampere’s law are:

Infinite straight lines, infinite planes, infinite solenoids, toroids.

Example 5.9 Find the m.f a distance s from a long

straight wire carrying a steady current I.

Solution. For s>R, let us choose a circular loop

of radius s around the wire. Ampere’;s law is

R I

s

enco IldB

By symmetry we see that B must be constant in magnitude and

tangent to the loop at every point on the loop. Then

IsBdlB o )2(s

IB o

2

Which agree with that of example 5.5 in the limit of long wire.

For a point inside the wire the circular

loop in this case has a radius sR. R

I s

To find the current enclosed by the loop, we note that

22 s

I

R

IJ enc

2

2

R

sIIenc

Ampere’s law now gives

2

2

)2(R

sIIsBdlB enco 22 R

IsB o

At s=0, the magnetic field is zero, as expected

encoId lB

Example 5.8 Find the m.f of a very

long solenoid, consisting of n turns

closely wound turns per unit length on

a cylinder of radius R and carrying a

steady current I.

Solution. The m.field for a long

solenoid is zero outside and uniform

inside. To calculate the internal

magnetic field, we can apply Ampere’s

law to the rectangular loop abcd of

length l and width w.

enco IldB

enco

a

d

d

c

c

b

b

a

IldBldBldBldB

For the portions bc and da B is to dl and for the portion cd

B=0, then only the 1st integral survives

enco

b

a

IldB

For the portion ab B is to dl and its magnitude is constant

along it

enco IBl

But the current crossing the surface enclosed by the loop is I multiplied by the No. of turns bounded by the loop

nIIl

NBNIBl ooo

With n is the No. of turns per unit length.

Example 5.10 Find the m.f of

a toroid of steady current I.

Solution From the symmetry, the

magnetic field lines form

concentric circles inside the

toroid and there is no magnetic

filed outside. This means that B

is constant in magnitude along

the Amperian loop and directed

tangent to it. Therefore, at a point

inside the coil we have

enco IldB

r

NIBNIrB o

o

22

Magnetic Vector Potential

0B

From Eq.(35) we have

)39(AB

d

rJB o

34

13

But

Now from Eq.(36) we have

drJB o 1

4

AfAfAf

Using

rJrJrJ

1

Zero since J is

independent on r.

)40(

44

d

rJd

rJB oo

Comparing Eq.(39) with Eq.(40) we get

)41(

44

ldId

rJA oo

Example 5.11 A spherical shell of

radius R, carrying a uniform charge

density , is set spinning at angular

velocity . Find the vector potential it

produces at point r.

Solution: The current density is given

by Eq.(29)

vd

dSv

d

dqvJ

Sdvd

rJA oo

44

cossinsincossincos0sin

ˆˆˆ

RRR

kji

rvBut

k

jiR ˆsinsinsin

ˆsincoscoscossinˆcossinsin

krrthatKnowing ˆ

kRjRiRrRrand ˆcosˆsinsinˆcossinˆ

cos222 RrrRrr

0sincos2

0

2

0

ddasNow

The x-component and thee z-component of A cancel and we get

022

2

0

3

cos2

sincossin

4ˆ

RrrR

ddRjA o

rRRrrRrRRrrRr

RjA o 2222

2sin

6ˆ

RrRrRrrR

rRthatKnowing

RrR

Rrr

r

RjA o

3

3

2 2

2sin6

ˆ

jrr

kji

rBut ˆsin00

cos0sin

ˆˆˆ

Rrrr

R

RrrR

Ao

o

3

4

3

3

If is along the z-axis and r is arbitrary we have

The magnetic field can be now found from Eq.(39)

ˆsinˆcossinˆsinsin

cossinsincossin00

ˆˆˆ

rjrir

rrr

kji

r

Rrr

R

RrrR

Ao

o

2

4sin

3

ˆsin3

Example 5.12 Find the vector potential for an

infinite solenoid of n turns per unit length,

radius R and current I.

Solution: It is difficult to use Eq.(41) since the

wire is infinite. Defining the m.flux as

)42(SdB

SdA

)43( ldA

Using Stoke’s theorem we get

Taking a closed circular loop inside the solenoid of radius s<R and

noting that A is constant along that loop and tangent to it we get

sAsB 22

ˆ

2ˆ

2s

nIs

BA o

in

B=onI

For outside we take a circular loop of radius s>R so we have

sARB 22

ˆ2

ˆ2

22

s

RnIR

s

BA o

out

Use Eq.(39) to check if it gives the correct values of B.

znI

snI

zs

zss

sAB o

o

inin ˆ

02

0

ˆˆˆ

1

2

0

02

0

ˆˆˆ

1

2

nIR

zs

zss

sAB

o

outout

Multipole Expansion of the Vector Potential

Consider a circuit of current I. The

vector potential at a distant point r>>r` is

ldIA o

4

0

cos11

nn

n

Pr

r

r

But rom Eq.(29) of Chapter 3 we have

0

1cos

1

4 nn

n

no ldPr

r

IA

For monopole (n=0) we have

0

01

4 n

omonopole ld

r

IA

For dipole (n=1) we have

ldrrr

Ildr

r

IA oo

dipole

ˆ

1

4cos

1

4 22

SdrldrrBut

ˆˆ

2

ˆ

4 r

rmA o

dipole

AISdImWith

Is the magnetic dipole moment with its direction is determined by

the R.H.R.