The induced emf is = Blv which produces a current I = /R where R is the resistance of the wire. There
is a force on the wire from the current and motion through the magnetic field of |Idl × B| = IlB, opposing the
motion of the wire. The equation of motion of the wire is then
m
2 2
2 2
FR
dv
B 2 l2
RF
=F −
v ⇒ v(t) = 2 2 + Ce−B l t/mR = 2 2 (1 − e−B l t/mR )
dt
R
B l
B l
since v(t = 0) = 0 ⇒ C = −RF/B 2 l2 .
(b) Calculate the current through the resistor R as a function of time. What is the direction of the current?
I(t) =
2 2
Blv(t)
F
=
(1 − e−B l t/mR ).
R
Bl
(World Scientific’s Problems and Solutions on Electromagnetism Problem 2054) A rectangle of
perfectly conductign wire having sides a and b, mass M and self-inductance L moves with an initial velocity v0 in
its plane, directed along its longest side, from a region of zero magnetic field into a region with a field B0 which
isuniform and perpendicular to the plane of the rectangle. Describe the motion of the rectangle as a function of
time.
B0 a
dI
d2 v
dv
2
√
+
ω
v
=
0
⇒
v
=
c
sin
ωt
+
c
cos
ωt
ω
=
.
= −B0 aI , L
= B0 av
⇒
m
1
2
dt
dt
dt2
mL
From the boundary conditions v(t = 0) = v0 ⇒ c2 = v0 ; I(t = 0) = 0 ⇒ c1 = 0 so v = v0 cos ωt. The position
(with s(t = 0) = 0) is then s(t) = vω0 sin ωt.
(World Scientific’s Problems and Solutions on Electromagnetism Problem 2070) A copper penny
is placed on edge in a vertical magnetic field B = 20 kGs. It is given a slight push to start it falling. Estimate
how long it takes to fall.
Assume the magnetic torque is always in equilibrium with the gravitational torque and that the potential
energy will be converted mainly into heat. Let θ be the angle between the plane of the penny and the vertical. A
ring of radii r and r + dr has a flux φ(θ) = πr2 B θ̇ cos θ passing through it. Then the induced emf and current are
dφ πr2 B θ̇ cos θ
= = πr2 B θ̇ cos θ ⇒ di =
=
dt
R
R
where R = 2πr/σhdr is the resistance of the ring that is h thick. The magnetic moment and torque follow from
the current:
di =
Brθ̇ cos θσhdr
πr2 B θ̇ cos θσhdr
πr2 B 2 θ̇ cos2 θσh
⇒ dm = πr2 di =
⇒ dτm = |dm × B| =
2
2
2
Let the radius of the penny be r0 so
Z
τm =
Z
r0
dτm =
0
πr4 B 2 θ̇ cos2 θσh
B 2 σr0 cos2 θ
πB 2 θ̇ cos2 θσh 3
r dr = 0
= τg = mgr0 sin θ = πr03 ρhg sin θ ⇒ dt =
·
dθ.
2
8
8gρ sin θ
Z
T =
Z
π/2
dt =
θ0
B 2 σr0
1
1 + cos θ0
B 2 σr0 cos2 θ
·
dθ =
− cos θ0 + ln
= 6.8 s.
8gρ
sin θ
8gρ
2
1 − cos θ0
(World Scientific’s Problems and Solutions on Electromagnetism Problem 2077) A magnetic dipole
m is moved from infinitely far away to a point on the axis of a fixed perfectly conducting (zero resistance) circular
loop of radius b and self-inductance L. In its final position the dipole is oriented along the loop axis and is at a
distance z from the center of the loop. Initially when the dipole is very far away, the current in the loop is zero.
4
(a) Calculate the current in the loop when the dipole is in its final position.
Z
Z
Z
dI
∂
= −L
B · dS
L[I(f ) − I(i)] = [B(f ) − B(i)] · dS ⇒ LI = B · dS
=−
dt
∂t
since initial conditions are I(i) = 0 and B(i) = 0 and denoting the final conditions I(f ) = I and B(f ) = B.
Consider a point ρeρ in the plane of the loop with radius vector r = ρeρ − zez . the integral is over the magnetic
induction at such a point due to m:
Z
Z Z µ0
µ0 3(m · r)r m
3(m · r)(r · ez ) m · ez
− 3 ⇒ B · dS =
−
ρdρdθ
B=
4π
r5
r
4π
r5
r3
Z b
3mz 2
m 3/2
µ0 m 2
µ0
− 2
ρdρ =
=
2π
[(b + z 2 )−1/2 − z 2 (b2 + z 2 )−3/2 ]
2
2
2
5/2
4π
(ρ + z
2
(ρ + z )
0
since m = mez and dS = ρdρdθez . The current is clockwise (by Lenz’s law) with a magnitude of I =
z 2 )−1/2 − z 2 (b2 + z 2 )−3/2 ].
(b) Calculate for the same positions the force between the dipole and the loop.
B0 = −
µ0 m
2
2L [(b
+
µ20 m
∂W
µ0 I
b2
b4
∂(m · B0 )
3µ20 m2 b4 z
e
=
−
e
⇒
F
=
−
.
=
−
=−
z
z
2
2
3
2
2
3/2
2 (b + z )
4L (b + z )
∂z
∂z
2L(b2 + z 2 )4
(World Scientific’s Problems and Solutions on Electromagnetism Problem 2079) A particle with
charge q is traveling with velocity v parallel to a wire with a uniform linear charge distribution λ per unit length.
The wire also arries a current I. What must the velocity be for the particle to travel in a straight line parallel to
the wire a distance r away?
Gauss’ flux theorem and Ampere’s circuital law with axial symmetry give
I
I
λ
µ0 I
λ
E · dS =
er
B · dl = µ0 I ⇒ B(r) =
eθ .
⇒ E(r) =
0
2π0 r
2πr
s
C
Let the particles velocity be v = vex and the total force acting on it should have no radial component:
F = Fe + Fm = qE + qv × B =
qµ0 I
qλ
qµ0 I
λ
λc2
qλ
er +
v(−er ) ⇒
−
v=0⇒v=
=
.
2π0 r
2πr
2π0 r
2πr
0 µ0 I
I
(World Scientific’s Problems and Solutions on Electromagnetism Problem 2086) A cylinder of
length L and radius R carries a uniform current I parallel to its axis.
(a) Find the direction and magnitude of the magnetic field everywhere inside teh cylinder. (Ignore end effects)
Ampere0 s law :
B=
µo Ir
eφ .
2πR2
(b) A beam of particles, each with momentum P parallel to the cylinder axis and each with positive charge q,
imponges on its end from the left. Show that after passing through teh cylinder teh particle beam is focused to
a point. (Make a thin lens approximation by assuming that the cylinder is much shorter than the focal length.
Neglect the slowing down and scattering of the beam particles by the material of the cylinder.) Compute the
focal length.
The magnetic force and resulting change in radial momentum for each particle is
Z
qvBL
µ0 qIL
F = qv × B = −qvBer ⇒ Pr = q vBdt =
=
r.
v
2πR2
The focal length is then Pr /P = r/d ⇒ d = P r/Pr =
2πR2 P
µ0 eIL
5
and is independent of r, as it should be.