Questions 2, 5, 7, 9, 12, 14, 15
Problems: 1, 3, 6, 7, 17, 20
Q2 A loop of wire is placed in a uniform magnetic field. For what orientation of the loop is the magnetic flux a maximum? For what orientation is the flux zero?
The magnetic flux is maximum when the magnetic field B is perpendicular to the loop of wire as in the diagram here:
The magnetic flux is minimal -- actually zero -- when the magnetic field B is parallel to the loop of wire as in the diagram here:
For illustration, it seemed better to draw this a nearly parallel rather than exactly parallel.
Q5 The bar in Figure 31.24 moves on rails to the right with a velocity v, and the uniform, constant magnetic field is directed out of the page. Why is the induced current clockwise? If the bar were moving to the left, what wold be the direction of the induced current?
There is a magnetic force due to motion through the magnetic field.
Fmag = q v x B
gives a vector downward for a positive charge (and a vector upward for a negative charge). Current is in the direction of the motion of a positive charge.
so this means the current flows down the moving bar, to the left along the bottom conductor, up through the resistor, and to the right along the top conductor. That is a clockwise flow through the entire circuit.
If the direction of the velocity is changed,
that changes everything. Now the vector v x B points upward. That is the direction of the magnetic force on a positive charge.
so that means the current flows up in the moving conductor. To complete its circuit, the current must flow to the left in the top conductor, down through the resistor, and to the right in the bottom conductor. This means a counterclockwise current throughout the entire circuit.
Q7 A large circular loop of wire lies in the horizontal plane. A bar magnet is dropped through the loop. If the axis of the magnet remains horizontal as it falls, describe the emf induced in the loop. How is the istuation altered if the axis of the magnet remains vertical as it falls?
When the bar magnet falls horizontally -- well, it falls vertically but it is lying in a horizontal position -- anyway, when it falls as shown here, the flux through the horizontal coil will be small. The flux is small because the magnetic field in nearly parallel to the coil and the direction of the magnetic field is such that when you consider the components perpendicular to the coil, there is about as much positive flux as negative flux. So, if the flux is small, the change in flux will also be small.
However, when the magnet is dropped as shown here, with its axis vertical, the flux through the coil is far greater -- so the change in flux will be greater, too.
Q9 Will dropping a magnet down a long copper tube produce a current in the tube?
Yes, this is an interesting and useful demonstration.
Q12 What happens when the speed at which the coil of a generator is rotated is increased?
Increasing the speed increases the rate at which the flux changes so that increases the emf or the voltage!
Q14 When the switch in Figure 31.25a is closed, a current is set up in the coil and the metal ring spring upward as in Figure 31.25b. Why?
The changing current in the coil causes a changing magnetic field in the iron core. This changing magnetic field causes an induced voltage and current in the metal ring. By Lenz's Law, this induced voltage or current will create a magnetic field in the direction opposite to the magnetic field in the iron core. These two fields repell each other!
Q15 Assume the battery in Figure 31.25a is replaced by an ac source and the switch is held closed. If held down, the metal ring on top of the solenoid becomes hot. Why?
The explanation above simply says "changing current" and "changing magnetic field". These can change because they are going from zero to some value or they can be changing because they are part of an alternating current (ac) circuit. The result is exactly the same.
If you continue to hold the ring in the alternating (changing) magnetic field, it will continue to have a current in it. Currents produce heat!
31.1 A 50-turn rectangular coil of dimensions 5.0 cm x 10.0 cm is dropped from a position where B = 0 to a new position where B = 0.50 T and is directed perpendicular to the plane of the coil. Calculate the resulting average emf induced in the coil if the displacement occurs in 0.25 s.
= - N d/dt = - N /t
= f - i
i = 0
f = [0.5 T] [(0.05 m)(0.10 m)] = 0.0025 T-m2
= f - i = 0.0025 T-m2
= - N d/dt = - N /t = - (50)(0.0025/0.25) V
= - 0.50 V
31.3 A powerful electromagnet has a field of 1.6 T and a cross-sectional area of 0.20 m2. If we place a coil having 200 turns and a total resistance of 20 ohms around the electromagnet and then turn off the power to the electromagnet in 20 ms (0.020 s), what is the current induced in the coil?
= - N d/dt = - N /t
= f - i
f = 0
i = [1.6 T] [0.20 m2] = 0.32 T-m2
= f - i = - 0.32 T-m2
= - N d/dt = - N /t = (200)(0.32/0.020) V
= 3200 V
31.6 A tightly wound circular coil has 50 turns, each of radius 0.10 m. A uniform magnetic field is turned on along a direction perpendicular to the plane of the coils. If the field increases linearly from 0 T to 0.6 T in 0.20 s, what emf is induced in the coil?
A = r2
A = (0.10 m)2 = 0.0314 m2
= A B
cos = cos 0 = 1
i = 0
f = (0.0314 m2)(0.6 T) = 0.01885 T-m2
= f - i = - 0.01885 T-m2
= - N d/dt = - N /t = (50)(0.01885/0.020) V
= 47 V
31.7 A 30-turn circular coil of radius 4 cm and resistance 1 ohm is placed in a magnetic field directed perpendicularly to the plane of the coil. The magnitude of the magnetic field varies in time according to the expression B = 0.010 t + 0.040 t2, where t is in seconds and B is in tesla. Calculate the induced emf in the coil at t = 5.0 s.
B = 0.010 t + 0.040 t2
dB/dt = 0.010 + (0.040)(2 t)
At t = 5.0 s,
dB/dt](5.0 s) = 0.010 + (0.040)(2 x 5)
dB/dt](5.0 s) = 0.010 + (0.040)(10)
dB/dt](5.0 s) = 0.010 + 0.40 = 0.41
A = r2
A = (0.04 m)2 = 0.005 m2
= A B
since cos = cos 0 = 1
= - N d/dt = - N d(AB)/dt = - NA dB/dt
= - (30)(0.005)(0.41) V = 0.062 V
While that is all the textbook asks for, since we also know the resistance of the coil, we might as well extend this just a bit and ask for the current induced in the coil as well.
V = IR
I = V/R
I = 0.062 V / 1
I = 0.062 A
31.17 A Boeing 747 jet with a wing span of 60.0 m is flying horizontally at a speed of 300 m/s over Phoenix where the direction of Earth's magnetic field is 58o below the horizontal. If the magnetiude of the magentic field is 50.0 T (microTeslas), what is the voltage generated between the wing tips?
A Boeing 747 jet with a wing span of 60.0 m is flying horizontally at a speed of 300 m/s over Phoenix where the direction of Earth's magnetic field is 58o below the horizontal. If the magnitude of the magentic field is 50.0 T (microTeslas), what is the voltage generated between the wing tips?
For the case of v perpendicular to B, we found
= - B l v
But it is the component of B that is perpendicular to v (or vice versa!) that is important, so the more general form of that equation is really
= - B l v sinwhere is the angle between B and v
= - B l v sin
= - B l v sin 58o
= - (50 x 10 - 6)(60)(300)(0.790) V
= - 0.711 V
31.20 In Figure P31.20, the bar magnet is moved toward the loop. Is Va - Vb positive, negative, or zero?
What do we have to work with? Lenz's Law tells us that the induced current will try to get the flux back to what it was before the change ("status quo ante").
Initially, with the magnet far away, the flux through the loop is zero.
As the magnet moves into the coil, the flux increases. Which direction is the magnetic field?
The magnetic field points from the North pole to the South pole. It may be a little easier at this point to redraw the magnetic field as a single vector.
Now we have changed from no flux to having a flux due to a magnetic field pointing "back" into the screen. According to Lenz's Law, the induced magnetic field will oppose this change.
The "induced magnetic field" is the magnetic field caused by the induced current.
Therefore, the induced current in the loop must be clockwise as we look at it. That means the current in the rest of the circuit must be as shown by the arrows in the sketch above.
All of this means that current flows through the resistor R from a to b. For that to happen, the voltage at a must be higher than the voltage at b.
Va > Vb
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(c) Doug Davis, 2002; all rights reserved