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Quiz 21 (ANSWERS ) CQ; Problems 1, 3, 4, 5, 6, 9, 10, 11, 12 (TEST 3)

TURN IN: CQ ; Problems 1, 3, 4, 5, 6, 9, 10, 11, 12

Problem Hints below:

error log #1: No errors reported

error log #2: No errors reported

I will look for important steps in your odd numbered solutions..

1. Use equation 21-1(a) with N = 1. In that case the magnitude of the emf |E| = |change in flux/ change in time|. Change in time and the flux at the beginning and end of the time interval are given; get the change in flux by subtracting them. .

3.

The induced magnetic field field opposes the change in the external field. The induced current causes the induced magnetic field in the loop. The external field increases in magnitude while pointing right in figure 21-47. Thus, the induced magnetic field points left.

That means the current flows up the vertical left wire . Use the right hand rule in chapter 20, page 557, or my version of the right hand rule. My version says wrap your fingers in the direction of the induced current flowing in the loop. Your thumb points in the direction of the induced magnetic field through the loop.

4. Use equation 21-1(a) with N = 1. In that case, the magnitude of the emf |E| = |change in flux/ change in time|. Change in time is given and the fluxes at the beginning and end of the time interval can be computed. Change in flux =
0 - B*A where B = 1.10 T and A is the area of a circle of diameter 0.096 m.

5. Use equation 21-1(a) with N = 1. In that case, the magnitude of the emf |E| = |change in flux/ change in time|. Change in time is given and the fluxes at the beginning and end of the time interval can be computed. Change in flux =
B*A*cos90 - B*A*cos0 where B = 1.5 T and A is the area of a circle of diameter 0.120 m.

6. Use equation 21-1(a) with N = 1. In that case, the magnitude of the emf |E| = |change in flux/ change in time|. Change in time is given and the fluxes at the beginning and end of the time interval can be computed. Change in flux =
B_f*A*cos180 - B_i*A*cos 0 where B_f = 0.25 T, B_i= 0.63 T and A is the area of a circle of diameter 0.102 m.

9. Here is a hint to fig 21-29 (a) . Get the direction of the external magnetic field though the loop. That field is due to the wire. To get the field direction, point your right thumb in the direction of the current in the wire. Use the right hand rule for long wires, Table 20-1, 1.

Given that rule, below the wire your right fingers point out of the page and through the loop . Thus the magnetic field due to the wire points out of the page and through the loop. Since I increases, that external field increases while pointing out. The induced magnetic field must thus point in to oppose the change in the external field from the wire.

The induced current causes the induced magnetic field in the loop. From Chapter 20 and my right hand rule, then what should be the direction of the induced current in the loop, clockwise or counter clockwise ? Use the right hand rule in chapter 20, page 557, or my version of the right hand rule discussed in problem 3.

10. See figure 20-30, which shows the field of a solenoid. The left side of the coil is a south pole and the right side of the coil is a north pole. Thus at the location of the loop in fig 21-50, the field points rightward in the picture. The field through the loop is decreasing since the coil is moving rightward.

To oppose that decrease, the induced magnetic field in the loop must also point rightward. The induced current causes the induced magnetic field in the loop. To get the direction of the induced current flowing in the loop use the right hand rule discussed in problem 3. Use the right hand rule in chapter 20, page 557, or my version of the right hand rule discussed in problem 3 .

Does the current flow up or down the side of the loop closest to you in figure 21-50?

11.

(a) Use equation 21-1(a) with N = 1. In that case, the magnitude of the emf |E| = |change in flux/ change in time|. Change in time is given and the fluxes at the beginning and end of the time interval can be computed. Change in flux =
B_f*A*cos180 - B_i*A*cos 0 where B_f = 0.45 T, B_i= 0.52 T and A is the area of a circle of given diameter

(b) Let positive be into the page, in the initial direction of the changing external magnetic field . By the right hand rule your fingers wrap clockwise in the direction of the current inducing a magnetic field to oppose the decrease in the magnitude of the changing external magnetic field.

12. (a) See figure 21-12. (b) E = vB from equation 21-4. We did not review this concept very well, but we will return to the topic later. In the rest frame of the charges, the velocity is zero. Thus the upward force causing the charge to move must not be magnetic but must be electric.

See discussion page 590, fig 21-12, which shows the electrons in the laboratory frame in which the electrons are moving. Now turn to page 591, bottom, which conveys the wisdom in the electron's rest frame. There, the particle has the same charge and mass. The upward force must be from another source, an electric field . The magnitude of this electric force is e*E, where e = electron charge magnitude and E = electric field magnitude. That force must be the same in either the laboratory or rest frame since in either frame, the net force = m*acceleration, where acceleration is invariant between frames. Thus force in rest frame = force in laboratory frame:

e*B = e*v*E.

This leads to B = v*.

More hints later