Lecture 16 :
Key words :
Ampere’s law, Field due to solenoid, current sheet,.
Problems :
1. Two long parallel wires carry 5 A current each in the opposite directions. The distance between the
wires is 0.6 m. Find the magnetic field at a distance of 0.4 m from the wire to the right.
P
0.6 m
0.4 m
2. Consider three long parallel wires whose intersection with the plane of the paper forms an
equilateral triangle of side √3 m. The wires carry 5 A current in the directions shown. Calculate the
field at the location of the upper wire by the other two. What is the force per unit length on the
upper wire?
3. A very large number of long parallel wires carry current I each. The separation between adjacent
wires is d and their current directions alternate. Find an expression for the field strength at the
location of the leftmost wire. What would be the field strength if the number of wires were infinite?
D d
4. A coaxial cable has an inner conductor of radius Rwhich is surrounded by an outer conductor of
inner radius 1.5R and outer radius 2R. a current I enters through the inner conductor and comes out
through the outer conductor. The current is uniformly distributed across the cross sections. Find the
magnetic field at a distance r from the axis.
5. A cylindrical conductor of radius a has an off axis hole of radius b running through the cylinder.
The distance between the axis of the cylinder and the hole is d. The cylinder carries a current I
uniformly distributed over the cross section. Determine the magnetic field at a point inside the hole
which is at a distance r from the axis of the cylinder.
6. A cylindrical conductor of radius a has a non uniformly distributed current. The current density
at a distance r from the axis of the cylinder is 𝐽 = 𝐽0 𝑟/𝑎. Calculate the magnetic field strength both
inside and outside the cylinder.
7. Consider an semi-infinite current sheet of finite width w and negligible thickness which carries
a current I along its length. Calculate the magnetic field at a distance d from one of the edges.
8. A solenoid of radius 10 cm and length 1.5 m, carries a current 20 A in its windings. If the field at
the centre of the solenoid is 100 Gauss, how many turns are there in the solenoid?
9. A toroid of inner radius 18 cm and outer radius 20 cm, carries a current of 10 A. What are the
maximum and minimum values of the field strength in the toroid, if it has 1000 turns?
10. A long cylinder of radius 2R has a coaxial hole of radius R. The cylinder carries a current I
distributed uniformly across its cross section. Calculate the field at a point on the axis.
Hints for solutions to Problem
1. Fields subtract as P is situated to the right side of oppositely directed currents. Since the outward
field is stronger, the net field is outward. 𝐵 =
𝜇0 𝐼 1
1
( − 1)
2𝜋 0.4
= 1.5 × 10−6 T.
2. Field due to each of the wires is perpendicular to the line joining the vertex to the wires. The net field
is along the base from left corner to right (1 to 2) as the perpendicular components cancel. (The
direction of the field due to the wires numbered 1 and 2 is indicated at the position of the top wire.
1
𝜇 𝐼
0
The field is 2𝜋𝑎
2 cos 300 = 10−6 T. The force on the top wire due to both wires is repulsive.
Resolving parallel and perpendicular to the base, the components parallel to the base cancel, leaving
𝜇 𝐼 𝐼
only the perpendicular components. Force per unit length is 𝐹 = 2 0 1 2 cos 300 = 5 × 10−6N.
2𝜋𝑎
𝜇 𝐼 1
1
1
1
Direct application of Ampere’s law gives 𝐵 = 0 ( − + − + ⋯ ). As the number of
2𝜋 𝑑
2𝑑
3𝑑
4𝑑
1
2
terms increase, the quantity within the bracket approaches 2𝑑 ln 2.
3.
4. For 𝑟 < 𝑅 , the problem is same as for a single wire with uniform distribution of current. The
𝜇 𝐼 𝑟
𝜇0 𝐼
field is 𝐵 = 2𝜋0 (𝑅2 ) .For 𝑅 < 𝑟 < 1.5𝑅, 𝐵 = 2𝜋𝑟
. For 1.5𝑅 < 𝑟 < 2𝑅 , the cross sectional area of
the outer cable is
𝜋 (𝑟 2 −
9𝑅2
).
4
7𝜋𝑅2
,
4
The cross sectional area of outer cable enclosed by the circle of radius r is
The total current enclosed by a circle of radius is then 𝐼 (1 −
𝜇 𝐼 16
9
4
7𝑅2
𝑟 2 −( )𝑅2
) . Using
4
4𝑟 2
0
this in Ampere’s law, 𝐵 = 2𝜋𝑅
( 7 − 7𝑅2 )For𝑟 > 2𝑅, the field is zero.
5.
Use the principle of superposition. Fill up the hole with equal and opposite current densities
𝐼
.
𝜋(𝑎 2 −𝑏2 )
𝐼𝑎 2
𝐼𝑏2
We then have a current 𝐼1 = 𝑎2 −𝑏2 in the original direction and 𝐼2 = 𝑎2 −𝑏2 . Use Ampere’s
law and show that the field inside the hole is independent of r.
6.
𝑎
𝜇 𝐼
0
Ampere’s law, 𝐵 = 2𝜋𝑟
⇒𝐵=
Thus for 𝑟 < 𝑎, 𝐵 =
7.
𝑟
𝑎
2𝜋𝐽0 𝑎 2
. Thus for 𝑟 > 𝑎, using
3
𝑟
𝑟
2𝜋𝜇0 𝐽0 𝑟 3
𝜇0 𝐼𝑒𝑛𝑐𝑙 = 𝜇0 ∫0 2𝜋𝑟 𝐽0 (𝑎) 𝑑𝑟 = 3𝑎
.
Total current through the conductor is 𝐼 = ∫0 𝐽0 ( ) 2𝜋𝑟 𝑑𝑟 =
𝜇0 𝐽0 𝑟 2
.
3𝑎
𝜇0 𝐽0 𝑎 2
.
3𝑟
For 𝑟 < 𝑎, 2𝜋𝑟 𝐵 =
Take a current sheet of width dx. The current in the strip is
𝐼𝑑𝑥
𝜇0 𝑥+𝑑.
Integrating, 𝐵 =
𝜇0 𝐼
𝑤+𝑑
ln 𝑑 .
𝑤
𝐼𝑑𝑥
.
𝑤
The field due to this strip is 𝑑𝐵 =
x
w
8. Since the ratio of length to diameter is large, the solenoid could essentially be considered as an
ideal one and edge effects may be neglected. Substituting the given values in 𝐵 = 𝜇0 𝑛 𝐼,
we get, 𝑛 = 398, which gives the number of turns 𝑁 = 597.
9. Ans. 0.011 T and 0.01 T
10. Ans. Zero (see problem 5)
Lecture 17 :
Key words :
Vector Potential, Gauge Invariance, Coulomb Gauge, Magnetic Flux, Emf, Faraday’s Law.
CORRECTIONS
1. In several places (indicated below) ⃗A has been printed as vecA. These have to be replaced.
2. Page 2 : line 2 : Equation : ∇φ = ⃗E, should read −∇φ = ⃗E. (a minus sign was missing in the equation)
⃗
3. Page 3 : three places line 1, line 2 and line 12 : vecAA
⃗ . Also remove a big horizontal line on this page.
4. Inside Exercise 1 : lines 3 and 4 : vecAA
5. Correct all example numbers.
⃗ , also on page 4 of the example line 4 from bottom : vecAA
⃗
6. Example 16 : Inside page 2 : vecAA
7. Example17 : inside : page 1, line 5 : Equation : insidesolenoid inside solenoid also next line a space
between 0 and outside.
⃗ in two places.
8. Inside example 17 : Page 2 line 3 : vecAA
⃗
9. Page 5 : Remove a big horizontal line. Line 5 from bottom vecAA
Problems :
⃗ = 5𝑥𝑖̂ + 3𝑦𝑗̂ + 𝑐𝑧𝑘̂, where c is a constant. Calculate the
1. The magnetic field B in a region is given by 𝐵
value of c.
⃗ = 𝑥 2 𝑦𝑧𝑖̂ + 𝑥𝑦 2 𝑧𝑗̂ − 2𝑥𝑦𝑧 2 𝑘̂. Obtain an
2. Magnetic field in a region is given by the expression 𝐵
expression for the vector potential corresponding to this field.
3. Obtain an expression for the vector potential corresponding to the magnetic field produced by a
current sheet in the xy plane carrying a linear current density K in the x direction.
4. Two long, parallel wires carrying current I in opposite directions are at a distance d from each other.
The currents run parallel to z axis, with one of the wires at x=0 while the other at x=d. Calculate the
vector potential at a point P in the xy plane at a position 𝑟⃗⃗ as shown.
1
X y
P
r

X x
5. A rectangular loop of sides𝑎 × 𝑏 lies in the xy plane. A uniform magnetic field B exists everywhere in
the region directed perpendicular to the plane of the loop. Calculate the emf generated when
a. The strength of the magnetic field is decreased from its maximum value B0 to zero
steadily over a time T.
b. The loop is moved with a speed v in the xy plane.
c. The loop is rotated about one of the sides with a uniform angular speed .
6. A circular coil of radius 40 cm lies in the xy plane in a region where the magnetic field is given by
⃗ = 0.2 cos(150𝜋𝑡)𝑖̂ + 0.4 sin(100𝜋𝑡) 𝑗̂ + 0.6 sin(50𝜋𝑡)𝑘̂
𝐵
Find the emf induced in the loop. If the loop has a resistance of 10 , what would be the rms value of
the induced current?
Hints for solutions to Problem
1. B is solenoidal, i.e. ∇ ∙ 𝐵 = 0 gives 𝑐 = −8.
2. Take 𝐴 = 𝑓1 (𝑥, 𝑦, 𝑧)𝑖̂ + 𝑓2 (𝑥, 𝑦, 𝑧)𝑗̂ + 𝑓3 (𝑥, 𝑦, 𝑧)𝑘̂. Use
𝜕
𝜕
3. (∇ × 𝐴)𝑥 = 𝜕𝑦 𝐴𝑧 − 𝜕𝑧 𝐴𝑦 =
𝜕𝑓3
𝜕𝑦
−
𝜕𝑓2
𝜕𝑧
= 𝑥 2 𝑦𝑧
This gives 𝑓3 = 𝑎𝑥 2 𝑦 2 𝑧, 𝑓2 = −𝑏𝑥 2 𝑦𝑧 2 with 𝑎 + 𝑏 = 1/2. (𝑓2 and 𝑓3 could also have an
additive constant which is a function of x alone but this will not be feasible because such
constants will not be consistent with the y and z components of the curl.) Taking y and z
components give us two pairs of relations, 𝑓1 = 𝑐𝑥𝑦 2 𝑧 2 , 𝑓3 = −𝑑 𝑥 2 𝑦 2 𝑧 with𝑐 + 𝑑 = 1/
2 and 𝑓2 = 𝑒𝑥 2 𝑦𝑧 2 , 𝑓1 = −𝑓 𝑥𝑦 2 𝑧 2 with 𝑒 + 𝑓 = −1. These equations are consistent if
1
1
𝑎 = 𝑑 = 0, 𝑏 = 𝑐 = 1/2, 𝑒 = 𝑓 = −1/2. Thus 𝑓1 = 2 𝑥𝑦 2 𝑧 2 , 𝑓2 = 2 𝑥 2 𝑦𝑧 2 , 𝑓3 = 0. Thus 𝐴 =
1
2
(𝑥𝑦 2 𝑧 2 𝑖̂ − 𝑥 2 𝑦𝑧 2 𝑗̂).
4. Use superposition principle. The vector potential is along z direction and is sum of two
vector potentials, one due to each wire. Using the expression for the vector potential
𝜇 𝐼
0
(ln(𝑟 − 𝑑 cos 𝜃) − ln 𝑟). For ≫ 𝑑, 𝐴𝑧 =
worked out in Example 1, we get 𝐴𝑧 = − 2𝜋
𝜇0 𝐼𝑑
2𝜋𝑟
cos 𝜃.
5. The normal to the loop is along the z direction. Thus the flux through the loop is Φ = 𝐵𝑎𝑏 .
a. If the magnetic field varies with time ℰ = −
b. Zero, as the flux does not change.
𝑑Φ
𝑑𝑡
=
𝐵0 𝑥𝑦
𝑇
.
c. As the plane rotates, the normal to the plane makes an angle 𝜃 = 𝜔𝑡 with the
magnetic field, so that Φ = 𝐵𝑎𝑏 cos 𝜃. The emf is ℰ = −
𝑑Φ
𝑑𝑡
= 𝐵𝑎𝑏𝜔 sin 𝜔𝑡.
6. Emf is given by ℰ = −47.3 cos(50𝜋𝑡). Maximum current is 4.73 A, rms current is 3.35 A.
Lecture 18 :
Key words :
Motional emf, Induced current, Lenz’s Law.
Problems :
1. A circular loop of radius 10 cm, having a resistance of 100 , lies on the plane of the paper. A
magnetic field B acts perpendicular to the plane of the loop and points inward. If the magnetic field
decreases at a uniform rate from its initial value 500 Gauss to a value of 50 Gauss in 25 ms, find the
current in the loop. Which direction does the current point as seen from above the plane of the
paper?
2. A rectangular conducting loop of dimensions 𝑎 × 𝑏 and resistance R is taken out of the magnetic
field B with a constant speed v, as shown. Determine the emf generated and the direction of the
current flow.
V
Vv b
a
V v
3. A circular loop of radius 20 cm and resistance 10  symmetrically surrounds a long solenoid having a
radius of 2.5 cm and 1600 turns per meter. The plane of the coil is perpendicular to the axis of the
solenoid. The current in the solenoid changes from a value of 10 A to 30 A in 2 seconds. Calculate
the maximum current induced in the circular coil.
4. A rectangular conducting loop of sides 50 cm x 30 cm lies in a magnetic field directed perpendicular
to its plane. A 1.5 V light bulb is in the loop, the resistance of rest of the loop can be neglected. At
what uniform rate should the magnetic field strength be increased so that the light bulb shines with
maximum brightness. What is the direction of the induced current in the loop?
Vv 30 cm
V 50 cm
5. A square loop of side a rotates such that the angle between the normal to the loop and a fixed
magnetic field B varies with time as 𝜃(𝑡) =
𝜃0 𝑡
.
𝑇
Find the maximum emf in the loop.
6. A conducting rod AB is moving with a speed of 0.1 m/s at right angles to its length, in contact with
two parallel rails of a circuit which are separated by 0.5 m. A uniform magnetic field of strength 1 T
is directed perpendicular to the plane of the circuit. The rod has a resistance of 0.2  while the
resistance of the other parts of the circuit may be neglected. How much work is needed to be done
to keep the rod moving with a constant speed? Show that this amount of work is dissipated as Joule
heat.
AA
A 0.5 m
AB
7. A conducting rod I of length a and resistance R s pivoted at a point in the region of a uniform
magnetic field B. The rod is free to rotate in a plane perpendicular to the direction of the magnetic
field B. If the rod rotates with a constant angular velocity w, find the emf between the two ends of
the rod. If the circuit is completed by connecting the pivot by resistanceless wire so that it is in the
shape of a sector of a circle a, what is the force and torque that acts on the rod?
8. A long straight wire carries a current I. A square loop of side a having a resistance R and lying in the
plane of the paper moves with a constant velocity v parallel to the wire. Calculate the emf induced in
the loop.
I
b
a
9. Consider the same arrangement as in Problem 8 but the square loop is moving perpendicular to the
wire. What is the emf in this case when the distance from the wire is b?
10. A conductingrodof resistance R slides frictionlessly along a pair of parallel conducting rails on a
table. The rails are separated by a length a and are connected by a fixed conducting section to
complete a circuit. The rails and the connecting section have negligible resistance. The rod is
connected to a mass by a massless string which passes over a pulley. A uniform magnetic field of
strength B is directed perpendicular to the surface of the table. Obtain an expression for the
instantaneous velocity v of the mass. What is the terminal speed?
mg
Hints for solutions to Problem
1. The time dependence of the field can be expressed by the equation 𝐵(𝑡) = 5 × 10−2 (1 −
36𝑡), where B is expressed in Tesla and t in seconds. The area of the loop is 𝜋𝑟 2 =
𝑑𝐵
0.0314 m2 . Thus the emfℰ = −𝜋𝑟 2 𝑑𝑡 = 0.057 V. The induced current is 0.57 mA. As the
magnetic flux is decreasing downwards, the induced current must be so as to increase it in
the downward direction. The current should thus be clockwise as viewed from above.
2. The instantaneous area of the loop which is inside the magnetic field is 𝐴(𝑡) = 𝑏(𝑎 − 𝑣𝑡).
The emf is ℰ = −
𝑑Φ
𝑑𝑡
𝑑𝐴
= −𝐵 𝑑𝑡 = 𝐵𝑏𝑣. Since the magnetic field is out of the page, theemf is
positive, implying the current is positive, i.e. it is in the counterclockwise direction. (convince
yourself that this is consistent with Lenz’s law.)
3. The field due to solenoid is confined within the solenoid and is given by 𝐵(𝑡) = 𝜇0 𝑛𝐼(𝑡). The
flux enclosed by the circular loop is Φ(𝑡) = 𝜋𝑟 2 𝜇0 𝑛𝐼(𝑡), where r is the radius of the
solenoid (not of the loop). emfℰ = −𝜋𝑟 2 𝜇0 𝑛
𝑑𝐼(𝑡)
𝑑𝑡
= −3.95 × 10−6 V. The current is 0.395
A.
4. For full brightness the emf in the circuit should be 1.5 V. dB/dt required for this is 10 T/s.
2
⃗ ∙ 𝑆 = 𝐵𝑎2 cos (𝜃0 𝑡). Emf is – 𝑑Φ = 𝐵𝑎 𝜃0 sin (𝜃0 𝑡).
5. The flux through the loop is Φ(𝑡) = 𝐵
𝑇
𝑑𝑡
𝑇
𝑇
Thus the maximum emf is 𝐵𝑎2 𝜃0 /𝑇.
6. The emf in the circuit is 𝐵𝑙𝑣 = 0.05 V, the current flowing in the circuit is 0.05/0.2=0.25 A.
The force acting on the rod is 𝐹 = 𝐼𝐵𝑙 = 0.125 N. The direction of the force is to the left,
opposite to the direction of the velocity of the rod. To keep the rod moving at constant
speed, the power required is 𝐹𝑣 = 0.0125 W. To see that this is the rate of Joule heat, note
that Joule heat is 𝐼 2 𝑅 = (0.25)2 × 0.2 = 0.0125 W.
7. Considering an imaginary sector of which the rod is a radial section, the flux through the
sector having an instantaneous angle  is Φ =
current in the circuit is 𝐼 =
acting is
𝑎𝐹
2
.
𝐵𝜔𝑎2
2𝑅
𝐵𝜃𝑎2
2
. Thus the emf developed is
. The force acting on the rod is 𝐼𝑎𝐵 =
𝐵2 𝑎3 𝜔
2𝑅
𝐵𝜔𝑎2
2
. The
. The torque
8. The flux enclosed does not change as the rectangle moves. Emf is zero.
9. Consider a strip of width dx which is located at a distance x from the current. The strip has
𝜇 𝐼
0
(𝑎𝑑𝑥). Suppose at a particular instant the
an area 𝑎𝑑𝑥. The flux through the strip is 2𝜋𝑥
nearer edge of the square is at a distance bfrom the current, the flux through the entire loop
is Φ(𝑡) =
ℰ=−
𝑎+𝑏 𝑎𝑑𝑥
∫
2𝜋 𝑏
𝑥
𝜇0 𝐼
𝜇0 𝐼𝑎
2𝜋
1
=
1 𝑑𝑏
𝜇0 𝐼𝑎
2𝜋
ln
𝑏+𝑎
𝑏
. The emf is obtained from Faraday’s law.
𝜇 𝐼𝑎2 𝑣
0
[𝑏+𝑎 − 𝑏] 𝑑𝑡 = 2𝜋𝑏(𝑏+𝑎)
I
x
a
b
dx
10. As the rod slides, a motional emf develops, resulting in a current 𝐼 =
provides a leftward force 𝐼𝑎𝐵 =
𝐵2 𝑎2 𝑣
𝑅
𝐵𝑎𝑣
𝑅
in the circuit. This
on the sliding road, which, in turn, provides an
upward tensile force on the mass. This acts in addition to the gravitational force mg acting
𝑑𝑣
downward. The equation of motion is 𝑚 𝑑𝑡 = 𝑚𝑔 −
𝐵𝑎2 𝑣
𝑅
. The differential equation can be
solved by direct integration. Assuming that the mass is released at 𝑡 = 0, the velocity is
𝑚𝑔𝑅
given by 𝑣(𝑡) = 𝐵2 𝑎2 (1 − 𝑒
−
𝐵2 𝑎2 𝑡
𝑚𝑅
). The terminal velocity can be found by taking 𝑡 → ∞ limit
of this expression, or more directly, by equating the net force to zero.
Lecture 19 :
Key words :
Induced Electric Field, Mutual Inductance, Self Inductance, Magnetic Energy
Problems :
1. A solenoid of length 1m and radius 10 cm has 5000 turns. The strength of the induced electric field at
a distance of 20 cm from the axis of the solenoid is 40 mV/m. Calculate the rate at which the current
through the coils of the solenoid is changing with time.
2. Two solenoids have radii R and 2R respectively. The number of turns in the former is N while the
number of turns in the latter is 2n per unit length. If the narrower solenoid is completely enclosed by
the other one, determine the mutual inductance of the pair.
2 R
2 2R
3. A solenoid of length 1 m and radius 5 cm has 5000 turns of wire. A circular coil of radius 1 cm having
100 turns is placed coaxially within the solenoid. Calculate (a) the mutual inductance of the pair and
(b) the induced electric field in the circular coil.
4. Current in a coil is changing at a rate of 150 A/s. If the emf developed in the coil is 0.5 V, what is the
self inductance of the coil?
5. A rectangular loop of dimensions a x b is at a distance d from a long straight wire. Calculate the
mutual inductance.
d
b
a
6. A conductor consists of two long thin conducting shells of radii a and b. A current I goes in through
the inner conductor and come out through the outer conductor. Calculate the magnetic energy per
unit length stored in the conductor and hence calculate the self inductance of the conductor.
7. A rod of length L is moving with a velocity v in a plane perpendicular to a uniform magnetic field.
What is the induced electric field at any point of the rod?:
8. A conducting rod OA of length L rotates with a constant angular speed  in the xy plane, about one
of its ends which is pivoted at the origin. A uniform magnetic field B points in the +z direction. Find
the induced electric field at any point on the rod and by taking the line integral of the electric field,
show that this leads to the value of emf developed, as calculated in Problem 7 of Lecture 18.
9. A cylindrical volume of radius R has a uniform axial magnetic field B which is increasing at a constant
rate . A conducting rod rests in the cylinder in a plane perpendicular to its axis, as shown in the
cross sectional view. Determine the electric field at any point of the rod, measuring the distance from
one end. By taking the line integral of the electric field, determine the potential difference between
the two ends of this rod and show that this is consistent with the emf calculation using Faraday’s law.
R B
R R
R L
10. A long cylindrical wire of radius R carries a current I, distributed uniformly over its cross section. Find
the magnetic energy density in the wire.
Hints for solutions to Problem
1. The field inside the solenoid is B(t) = m0 nI (t) . The flux enclosed by a circle of radius is
I
R 2 I
(t )   R B (t )  0n  R I (t ) Emf   0n  R
 2 rE  E   0n
.
t
2r t
Substituting data, the current changes at 255 A/s.
2
2
2
2. Consider a current flowing through the inner conductor. The entire flux due to this is
captured by the outer conductor. The field due to the inner solenoid is
𝜇0 𝑁𝐼
𝑙
inner solenoid only. The flux captured by 2n turns of the outer solenoid is
mutual inductance is
𝜇0
confined to the
𝜇0 𝜋𝑅 2 2𝑁𝑛𝐼
𝑙
. The
𝜋𝑅 2 2𝑁𝑛
𝑙
.
3. The flux trapped by the circular coil is 𝐵𝜋𝑟 2 = 𝜇0 𝑛𝐼 𝜋𝑟 2 𝑁𝑐𝑖𝑟𝑐𝑙𝑒 = 1.97 × 10−4 𝐼. The mutual
inductance is 1.97 × 10−4 H.
4. Ans. 3.33 mH.
5. If the current flowing through the wire is I, the flux trapped by the rectangle is
𝑑+𝑏 𝑎𝑑𝑥
𝜇0 𝐼 ∫𝑑
2𝜋𝑥
=
𝜇0 𝐼𝑎
2𝜋
ln
𝑑+𝑏
𝑑
. The mutual inductance is given by dividing this expression by
current I.
6. Using Ampere’s law one can see that non-zero magnetic field only exists in the region
𝜇 𝐼
0
between the two conducting shells and is given by 𝐵(𝑟) = 2𝜋𝑟
. In this region consider a
cylindrical shell lying between r and r+dr. The energy per unit volume being
contained in a shell of length lis
𝐵(𝑟)2
2𝜇0
1
𝐵(𝑟)2
2𝜇0
, the energy
2𝜋𝑟𝑙 𝑑𝑟 . Total energy is obtained by integrating this
𝑏
1
expression from a to b, giving 𝑈 = 4𝜋 𝜇0 𝐼 2 𝑙 ln 𝑎. Equating this to 2 𝐿𝐼 2 , one gets the self
inductance per unit length of the conductor.
7. The motional emf developed in the rod is 𝑣𝐵𝐿. Since the rod is symmetrical, the electric field
must be constant over its length. As emf is the line integral of the electric field, the field is
𝑣𝐵.
8. Consider a strip of length dr at a distance r from the pivot. The velocity of this strip is 𝑣 =
⃗ = 𝜔𝑟𝐵(𝜃̂ × 𝑘̂) = 𝜔𝑟𝐵𝑟̂ . The emf is
𝜔𝑟 𝜃̂ . Thus the electric field at this point is 𝐸⃗ = 𝑣 × 𝐵
𝐿
∫ 𝐸⃗ ∙ 𝑑𝑟 = 𝜔𝐵 ∫0 𝑟𝑑𝑟 = 𝜔𝐵𝐿2 /2.
X
y
X
X
X
𝜃̂
X dr
r

X
x
9. The electric field, by symmetry has a magnitude which depends on the distance r from the
centre and by Lenz’s law is directed tangentially to a circle of radius r, pointing anticlockwise.
R B
Rr Rh
R𝐸⃗
𝑟 𝑑𝐵
The electric field at a distance r from the centre of the circle is 2 𝑑𝑡 =
𝑟𝛽
2
. Taking the
𝐿 𝑟𝛽 ℎ
ℎ𝛽
component along the rod, ∫ 𝐸⃗ ∙ 𝑑𝑙 = ∫0 2 𝑟 𝑑𝑥 = 2 𝐿. This can be seen to be consistent
with Faraday’s law by drawing imaginary lines connecting the two ends of the rod to the
centre and calculating the flux change through the triangular region. The area of the triangle
is Lh/2. The rate of change of flux through this region is Lh/2.
10. Consider the region of the cylinder lying between r and r+dr. The energy stored in this
region is 𝑑𝑈 =
𝑅 𝜇0 𝐼 2 𝑙 3
𝑟 𝑑𝑟
4𝜋𝑅 4
∫0
=
𝐵(𝑟)2
2𝜇0
𝜇0 𝐼 2 𝑙
16𝜋
𝑟
2𝜋𝑟 𝑙 𝑑𝑟 where 𝐵(𝑟) = 𝜇0 𝐼 2𝜋𝑅2 . The total energy is thus given by 𝑈 =
.
Lecture 20 :
Key words :
Paramagnetism, Diamagnetism, Magnetization, Bound Current, Free Current, H-Field, Magnetic
Permeability, Magnetic Susceptibility, Hysteresis, Displacement Current, Maxwell’s Equations,
Electromagnetic Wave, Polarization, Wave Equation, Electromagnetic Spectrum, Poynting Vector,
Radiation Pressure.
Problems :
⃗ =
1. The magnetic field in a system having cylindrical symmetry is given by the expression 𝐵
𝜇0 𝐼
̂
2 𝜌 𝜃 , where  is the distance from the z axis and R is a constant having the dimension of
2𝜋𝑅
length. Find the current density that produces this field.
2. The magnetic field in the region between the plates of a parallel plate capacitor of circular
𝑡
𝜌
⃗ = 𝐵0 𝑒 −𝑅𝐶 𝜃̂, where R is the circuit
cross section of radius a varies with space and time as 𝐵
𝑎
resistance and C is the capacitance. Obtain an expression for the displacement current
between the plates.
⃗ =
3. Magnetic field intensity of a travelling wave at a point in free space is given by 𝐵
⃗ at that
𝐵0 sin(𝑘𝑧 − 𝜔𝑡) 𝑗̂. Find the displacement current density and the electric field 𝐷
point.
⃗⃗ parallel to its axis. Calculate
4. An infinitely long cylindrical wire has a uniform magnetization 𝑀
the bound current in the wire.
5. An infinitely long cylinder of radius R has a “frozen in” magnetization parallel to its axis,
⃗⃗ = 𝐴𝜌2 𝑧̂ , where r is the distance from the axis of the cylinder. Calculate the
given by 𝑀
bound current densities both in the volume and on the surface and show that the net bound
current is zero.
⃗⃗ =
6. Repeat the above problem when the frozen-in magnetization is directed azimuthally, i.e. when 𝑀
𝑘𝜌2 𝜃̂.
7. A monochromatic laser beam has uniform intensity over a circular area of radius 0.2 mm. If
the total power of the beam is 96 mW, find the magnitude of the magnetic field in the beam.
If the circular area totally reflects the light, what is the radiation pressure on the area?
8. The electric field for an elliptically polarized is given by the expression
𝐸⃗ = 𝐸0𝑥 cos(𝑘𝑧 − 𝜔𝑡) 𝑖̂̂ + 𝐸0𝑦 cos(𝑘𝑧 − 𝜔𝑡 + 𝛼) 𝑗̂
Where 𝐸𝑜𝑥 ≠ 𝐸0𝑦 and  is a nonzero constant. Find the magnetic field corresponding to the
above electric field and also the Poynting vector.
9. A plane electromagnetic wave travelling in vacuum is given by the expression 𝐸⃗ =
𝑥
5 cos (2𝜋 × 108 (𝑡 − 𝑐 )) 𝑗̂ V/m. Find the wavelength, the associated magnetic field, average
power density and the direction of power flow.
10. A point source of plane electromagnetic wave having a wavelength of 3 m, has a power
output of 750 W. Find the radiation pressure on a surface at a distance of 10 cm from the
source assuming that the surface on which the radiation falls is perfectly reflecting.
11. A long wire of cross sectional radius R is made of a material of susceptibility . The wire has a
⃗ ,𝑀
⃗⃗ and 𝐻
⃗ at a distance r
uniformly distributed current over its cross section. Find the fields 𝐵
from the axis of the wire. Also determine the bound currents in the wire.
12. An infinite solenoid carrying a current in its turns is filled with a material of susceptibility .
⃗ ,𝑀
⃗⃗ and 𝐻
⃗ inside the solenoid.
Find the fields 𝐵
Hints for solutions to Problem
1. We need to calculate the curl in the cylindrical coordinates (, , z), in which the curl is given
by
⃗ =(
𝛁 × ⃗𝑩
∂Bρ ∂Bz
1 ∂Bz ∂Bθ
1 ∂(ρBθ ) ∂Bρ
−
) 𝜌̂ + (
−
) 𝜃̂ + (
−
) 𝑧̂
ρ ∂θ
∂z
∂z
∂ρ
ρ
∂ρ
∂θ
Since B only has  component which depends only on , only the last term gives non-zero
contribution.
⃗ =
𝛁 × ⃗𝑩
1 ∂(ρBθ ) ∂Bρ
1 𝜇0 𝐼 𝜕 2
𝜇0 𝐼
(𝜌 )𝑧̂ =
(
−
) 𝑧̂ =
𝑧̂ = 𝜇0 𝐽
2
ρ
∂ρ
∂θ
𝜌 2𝜋𝑅 𝜕𝜌
𝜋𝑅 2
The current density corresponds to that of a long wire with a uniformly distributed current.
2. Calculating the curl of the magnetic field in the cylindrical coordinates, we get
1 ∂(ρBθ )
2𝐵0 − 𝑡
𝜕𝐸⃗
⃗⃗ = (
𝛁×𝑩
) 𝑧̂ =
𝑒 𝑅𝐶 𝑧̂ = 𝜇0 𝜖0
ρ
∂ρ
𝑎
𝜕𝑡
𝜕𝐸⃗
𝑡
2𝐵
The displacement current density is 𝜖0 𝜕𝑡 = 𝑎𝜇0 𝑒 −𝑅𝐶 𝑧̂
0
⃗ =−
3. Calculate curl in Cartesian coordinates. 𝛁 × ⃗𝑩
𝜕𝐸⃗
𝜕
𝜕𝑧
𝐵𝑦 𝑖̂ = −𝐵0 𝑘 cos(𝑘𝑧 − 𝜔𝑡)𝑖̂. Thus the
𝐵 𝑘
displacement current density is 𝜖0 𝜕𝑡 = −( 𝜇0 ) cos(𝑘𝑧 − 𝜔𝑡)𝑖̂. This can be used to deduce
0
𝐵0 𝑘
⃗ = 𝜖0 𝐸⃗ = ( ) sin (𝑘𝑧 − 𝜔𝑡)𝑖̂.
the electric field 𝐷
𝜇
0
4. Since the magnetization is constant, the bound volume current density is zero. There is,
⃗⃗ × 𝑛̂ = 𝑀
⃗⃗ × 𝜌̂ = 𝑀𝜃̂
however, a surface current density ⃗⃗⃗⃗
𝐾𝑏 = 𝑀
⃗ ×𝑀
⃗⃗ = −
5. The bound volume current is 𝐽⃗⃗⃗𝑏 = ∇
𝜕𝑀𝑧
𝜕𝜌
⃗⃗ ×
= −2𝑘𝜌𝜃̂ and bound surface current ⃗⃗⃗⃗
𝐾𝑏 = 𝑀
𝑛̂ = 𝑘𝑅 2 𝜃̂. Since the current densities are in the 𝜃̂ direction, we can determine the total current by
considering a rectangular sheet of length L containing the z-axis and the outward normal 𝑛̂. If we
consider a strip of width d, the volume current through the strip is |𝐽⃗⃗⃗𝑏 |𝐿𝑑𝜌. Thus the net volume
𝑅
current through one half of the cylinder of length L is ∫0 −2𝑘𝜌𝐿𝑑𝜌 = −𝑘𝐿𝑅 2 . The surface current is
|𝐾𝑏 |𝐿 = 𝑘𝐿𝑅 2 . Thus the net bound current is zero.
To find the magnetic field consider an Amperian loop of width L
but extending from a distance ρ to the right of the axis to far away
to the right. The field far away is zero, the contribution from the top
and bottom of the loop is also zero. Thus the line integral of B is BL.
𝐽⃗⃗⃗𝑏
⃗⃗⃗⃗
𝐾𝑏
𝑅
The amount of current enclosed is 𝑘𝑅 2 𝐿 − 2𝑘 ∫𝜌 𝜌𝐿 𝑑𝜌 = 𝑘𝜌2 𝐿.
⃗ = 𝜇0 𝑘𝜌2 𝑧̂ . The field outside is zero.
Thus 𝐵
One can also do this problem by observing that the H-field is zero
⃗⃗ = 𝑘𝜌2 𝑧̂ . Thus 𝐵 = 𝜇0 𝑀 = 𝜇0 𝑘𝜌2 𝑧̂ .
Inside the conductor the magnetization is 𝑀
Outside, all the three quantities, B, H and M are zero.
1 𝜕
⃗⃗ =
⃗⃗ × 𝑛̂ = 𝑘𝑅 2 𝜃̂ × 𝜌̂ = −𝑘𝑅 2 𝑧̂ . Since the
(𝑘𝜌3 )𝑧̂ = 3𝑘𝜌𝑧̂ and ⃗⃗⃗⃗
6. In this case 𝐽⃗⃗⃗𝑏 = ⃗∇ × 𝑀
𝐾𝑏 = 𝑀
𝜌 𝜕𝜌
current density is in the z-direction, we can determine the total current by taking a circular disk of
radius R perpendicular to the axis. To determine the total volume current consider an annulus of area
lying between 𝜌 and 𝜌 + 𝑑𝜌. The current through this annulus is 3𝑘𝜌(2𝜋𝜌𝑑𝜌). Integrating this from
0 to R gives the total volume current to be 2𝜋𝑘𝑅 3 . The surface current is obtained by multiplying the
surface current density with the circumference of the disk. Fields can be determined similar to earlier
problem
96×10−3
24
7. The mean Poynting vector is the power per unit area, which is 𝜋(2×10−4 )2 = 𝜋×10−5 . Equate
𝑐𝐵2
3×108 𝐵2
this to 〈𝑆〉 = 2𝜇 = 8𝜋×10−7 , we get, 𝐵 = 8 × 10−5 T.
0
𝐸⃗
8. Since the propagation is along the positive z direction, the magnetic field is given by 𝑘̂ × 𝑐 =
𝐸0𝑥
𝑐
2
𝐸0𝑥
𝑐𝜇
cos(𝑘𝑧 − 𝜔𝑡)𝑗̂ −
𝐸0𝑦
𝑐
cos 2 (𝑘𝑧 − 𝜔𝑡)𝑘̂ +
⃗
𝐵
cos(𝑘𝑧 − 𝜔𝑡 + 𝛼)𝑖̂. The Poynting vector is given by 𝐸⃗ × 𝜇 =
0
2
𝐸0𝑦
𝑐𝜇
cos 2 (𝑘𝑧 − 𝜔𝑡 + 𝛼)𝑘̂. As expected the energy flow is in the z
direction.
9. In this case 𝜔 = 2𝜋 × 108 = 𝑐𝑘 ⇒ 𝑘 =
2𝜋
𝜆
=
2𝜋
3
⇒ 𝜆 = 3 m. Direction of propagation is x
⃗ = (5) cos (2𝜋 ×
direction. The magnetic field is thus along the z-direction and is given by 𝐵
𝑐
𝑥
𝑥
108 (𝑡 − 𝑐 )) 𝑘̂ = 1.67 × 10−8 cos (2𝜋 × 108 (𝑡 − 𝑐 )) 𝑘̂ (T). The power density is the
⃗
𝐵
𝑥
Poynting vector 𝐸⃗ × 𝜇 = 0.067 cos2 (2𝜋 × 108 (𝑡 − 𝑐 )) 𝑖̂. Average power density is 0.033
0
W/m2.
10. A point source emits uniformly in all directions. The magnitude of power density at a
distance r from the source is the power falling per unit area of a sphere of radius r and is the
750
W
Poynting vector. 〈𝑆〉 = 4𝜋(0.1)2 = 5968.3 m2. The pressure (the factor of 2 is due to
reflection is
〈𝑆〉
𝑐
N
= 9.95 × 10−6 m2 .
⃗ = 𝐼𝑟 2 𝜃̂. Using this,
11. Using Ampere’s law for free current, the H field inside the wire is 𝐻
2𝜋𝑅
⃗ = 𝜇0 (1 + 𝜒)𝐻
⃗ = 𝜇0 𝐻
⃗ + 𝜇0 𝑀
⃗⃗ , which gives 𝑀
⃗⃗ = 𝜒 𝐼𝑟 2 𝜃̂̂ . Bound currents
inside the wire 𝐵
2𝜋𝑅
𝐼
⃗ =
can be obtained from the magnetization density following Problem 1, ⃗⃗⃗
𝐽𝑏 = 𝜒 𝜋𝑅2 𝑧̂ and 𝐾
𝐼
−𝜒 2𝜋𝑅 𝑧̂ .
⃗ = 𝑛𝐼𝑧̂ , 𝑀
⃗⃗ = 𝜒𝐻
⃗, 𝐵
⃗ = 𝜇0 (1 + 𝜒) 𝑛𝐼𝑧̂
12. Answers : 𝐻