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1. The conducting bar illustrated in the figure moves on
two frictionless, parallel rails in the presence of a
uniform magnetic field directed into the page. The bar
has mass m, and its length is l. The bar is given an
initial velocity vi to the right and is released at t = 0.
(a) Using Newton’s laws, find the velocity of the bar
as a function of time.
(b) Show that the same result is found by using an energy approach.
2. A conducting bar of length l rotates with a constant angular
speed  about a pivot at one end. A uniform magnetic field
B is directed perpendicular to the plane of rotation as shown
in the figure. Find the motional emf induced between the
ends of the bar.
3. A long solenoid of radius R has n turns of wire per unit
length and carries a time-varying current that varies
sinusoidally as I = Imax cos  t, where Imax is the maximum
current and  is the angular frequency of the alternating
current source.
(a) Determine the magnitude of the induced electric field
outside the solenoid at a distance r > R from its long
central axis.
(b) What is the magnitude of the induced electric field inside the solenoid, a distance r
from its axis?
4. A coil of 15 turns and radius 10.0 cm surrounds a long
solenoid of radius 2.00 cm and 1.00×103 turns/meter. The
current in the solenoid changes as I = (5.00 A) sin (120 t).
Find the induced emf in the 15-turn coil as a function of
time.
5. A loop of wire in the shape of a rectangle of width w and
length L and a long, straight wire carrying a current I lie on
a tabletop as shown in the figure. (a) Determine the
magnetic flux through the loop due to the current I. (b)
Suppose the current is changing with time according to I =
a + bt, where a and b are constants. Determine the emf that is induced in the loop if b =
10.0 A/s, h = 1.00 cm, w = 10.0 cm, and L = 100 cm. What is the direction of the induced
current in the rectangle?
6. The rotating loop in an AC generator is a square 10.0 cm on a side. It is rotated at 60.0 Hz
in a uniform field of 0.800 T. Calculate (a) the flux through the loop as a function of time,
(b) the emf induced in the loop, (c) the current induced in the loop for a loop resistance of
1.00 Ω, (d) the power delivered to the loop, and (e) the torque that must be exerted to
rotate the loop.
7. A conducting rectangular loop of mass M, resistance R, and
dimensions w by ℓ falls from rest into a magnetic field B as
shown in the figure. During the time interval before the top
edge of the loop reaches the field, the loop approaches a
terminal speed vT.
MgR
B 2 w2 .
(b) Why is vT proportional to R?
(a) Show that vT 
(c) Why is it inversely proportional to B2?
8. The figure is a graph of the induced emf versus time for a coil
of N turns rotating with angular speed ω in a uniform magnetic
field directed perpendicular to the axis of rotation of the coil.
What If? Copy this sketch (on a larger scale), and on the same
set of axes show the graph of emf versus t (a) if the number of
turns in the coil is doubled; (b) if instead the angular speed is
doubled; and (c) if the angular speed is doubled while the
number of turns in the coil is halved.
9. A conducting rod moves with a constant velocity v in a
direction perpendicular to a long, straight wire carrying a
current I as shown in Figure P31.58. Show that the
magnitude of the emf generated between the ends of the rod
is


 0 vI 
2r
In this case, note that the emf decreases with increasing r, as you might expect.
(Faraday’s law)
1. v = vie-t/ ,  = mR/B2l2
2.
1
B  l2
2
3. (a)
 0 nI max R 2
sin  t (b)
2r
4. –14.2 cos (120 t) mV
 0 nI max 
2
rsin  t
 0 IL h  w
ln(
) (b) –4.80 V, counterclockwise
2
h
6. (a) (8.00 mTm2) cos (377 t) (b) (3.02 V) sin (377 t) (c) (3.02 A) sin (377 t)
(d) (9.10 W) sin2 (377 t) (e) (24.1 mNm) sin2 (377 t)
7. (b) The emf is directly proportional to vT, but the current is inversely proportional to R. A
large R means a small current at a given speed, so the loop must travel faster to get FB =
mg.
(c) At a given speed, the current is directly proportional to the magnetic field. But the
force is proportional to the product of the current and the field. For a small B, the speed
must increase to compensate for both the small B and also the current, so vT B2.
8. (a) Amplitude doubles: period unchanged (b) doubles the amplitude: cuts the period in
half (c) Amplitude unchanged: cuts the period in half
5. (a)