Worksheet 6
ID
name
1A Figures shows a bar magnet is about to fall through a metal ring. The magnet is
still above the ring.
(a) Find the direction of the induced magnetic field.
(b) Find the direction of the induced current.
(c) Find the direction of the force exerted on the manget by the ring.
(a) The original magnetic field from the magnet is downward inside
the ring. Since it is increasing, the direction of the induced
magnetic field is opposite to that of the original magnetic field
as shown in Figure. Hence, the direction of induced magnetic
field in the ring is upward (out of paper in the top view).
(b) Using RHR-2, the current flows in counterclockwise as shown in
the top view.
(c) Since the induced magnetic field is against the original magnetic
field, the ring and the magnet repel each other. Hence, an upward force is exerted on the magnet by the ring.
1B Figure shows that the magnet is falling away from the ring after passing through
it.
(a) Find the direction of the induced magnetic field.
(b) Find the direction of the induced current.
(c) Find the direction of the force exerted on the manget by the ring.
(a) The original magnetic field from the magnet is downward inside the ring. Since it is decreasing, the direction of the induced
magnetic field is in the same direction as that of the original
magnetic field as shown in Figure. Hence, the direction of induced magnetic field in the ring is upward (out of paper in the
top view).
(b) Using RHR-2, the current flows in clockwise as shown in the top
view.
(c) Since the induced magnetic field is in the same direction as that
of the original magnetic field, the ring and the magnet attract
each other. Hence, an upward force is exerted on the magnet by
the ring.
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2. A conducting rod is free to slide along a pair of conducting rails,
in a region where a uniform and constant (in time) magnetic field
is directed into the plane of the paper, as the drawing illustrates.
(a) When the switch is open, the rod is forced to move to the
right by a hand. Find the direction of force exerted on the
rod by the magnetic field.
(b) Initially the rod is at rest, Describe the rod’s motion after
the switch is closed. Be sure to account for the effect of any
motional emf that develops.
(c) Suppose that the voltage of the battery in the circuit is 3.0 V ,
the magnitude of the magnetic field (directed perpendicularly
into the plane of the paper) is 0.60 T , and the length of the
rod between the rails is 0.20 m. Assuming that the rails are
very long and have negligible resistance, find the maximum
speed attained by the rod after the switch is closed.
(a) When the switch is open, a continuous current does not flow in the rod. However, there is a transient
current in the upward direction. In turn this transient current exerts a transient force on the rod.
The direction of the force is opposite to the direction of the velocity.
(b) When the switch is closed, a conventional current will flow along the conducting rails from the
positive toward the negative terminal of the battery. Since the rod is a conducting rod, current will
flow through the rod, from top to bottom. According to RHR-1, there will be a force that points to
the right on the conducting rod due to the magnetic field; therefore, the rod will be pushed and
accelerate to the right. As the rod moves to the right, the area bound by the ”loop” increases,
thereby increasing the magnetic flux through the loop. As the magnetic flux increases, an induced
emf appears around the ”loop.” According to Lenz’s law, the induced emf that appears will appear in
such a way so as to oppose the increase in the magnetic flux. This will occur if the induced emf
opposes the battery emf, with the result that the current in the rod begins to decrease and reaches
zero when the induced emf exactly offsets the battery emf. With no current in the rod, there is no
longer a magnetic force applied to the rod. With no force, there is zero acceleration. In other words,
from this point on, the rod moves with a constant velocity.
(c) The moving rod produces an emf given by
emf = vBL
When this emf reaches the voltage of the battery, the current vanishes. Hence,
vBL = V
→
v=
V
3.0 V
=
= 25 m/s .
BL
(0.60 T )(0.20 m)
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3. A flat coil of wire has an area A, N turns, and a resistance R. It is situated in a magnetic field such
that the normal to the coil is parallel to the magnetic field. The coil is then rotated through an angle of
90◦ , so that the normal becomes perpendicular to the magnetic field.
(a) Why is an emf induced in the coil?
(b) What determines the amount of induced current in the coil?
(c) How is the amount of charge that flows related to the induced current I and the time interval during
which the coil rotates?
(d) The coil has an area of 1.5 × 10−3 m2 , 50 turns, and a resistance of 140 Ω. During the time when it is
rotating, a charge of 8.5 × 10−5 C flows in the coil. What is the magnitude of the magnetic field?
(a) An emf is induced in the coil because the magnetic flux through the coil is changing in time. The flux
is changing because the angle φ between the normal to the coil and the magnetic field is changing.
(b) The amount of induced current is equal to the induced emf divided by the resistance of the coil (see
Equation 20.2).
(c) According to Equation 20.1, the amount of charge ∆q that flows is equal to the induced current I
multiplied by the time interval ∆t = tt0 during which the coil rotates, or ∆q = I(tt0 ).
(d) According to Equation 20.1, the amount of charge that flows is ∆q = I∆t. The current is related to
the emf in the coil and the resistance R by Equation 20.2 as I = emf/R. The amount of charge that
flows can, therefore, be written as ∆q = emf∆t/R. The emf is given by Faradays law of
electromagnetic induction as
¶
µ
¶
µ
BA cos φ − BA cos φ0
∆Φ
= −N
emf = −N
∆t
∆t
where we have also used Equation 22.2, which gives the definition of magnetic flux as Φ = BA cos φ.
With this emf, the expression for the amount of charge becomes
³
´

cos φ0
−N BA cos φ−BA
∆t
 ∆t = −N BA(cos φ − cos φ0 )
∆q = 
R
R
Solving for the magnitude of the magnetic field yields
B=
−(140 Ω)(8.5 × 10−5 C)
R∆q
=
= 0.16 T
N A(cos φ − cos φ0 )
(50)(1.5 × 10−3 m2 )(cos 90◦ − cos 0◦ )
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