In the diagram at right, switch S is initially closed, and the battery is driving a current counter-clockwise through the top loop. Suppose that the switch is opened at time t1, help open until time t2, and then closed again. Describe the current that will be induced in the bottom loop while all this occurs. O There will be a brief pulse of counter-clockwise current, both when S is opened and when it is closed. O There will be a brief pulse of counter-clockwise current when Sis opened, and then a brief pulse of clockwise current when S is closed. O There will be a steady clockwise current before S is opened, a steady counter-clockwise current while S remains open, and then a steady clockwise current again after S is closed. O There will be a brief pulse of clockwise current, both when S is opened and when it is closed. O There will be a brief pulse of clockwise current when S is opened, and then a brief pulse of counter-clockwise current when S is closed. O There will be no current before S is opened, a steady counter-clockwise current while S is open, and then no current again after S is closed.

Physics for Scientists and Engineers: Foundations and Connections
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Author:Katz, Debora M.
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Chapter32: Faraday’s Law Of Induction
Section: Chapter Questions
Problem 30PQ: Two circular conducting loops labeled A and B are close together and their axes are parallel. Loop A...
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In the diagram at right, switch S is initially closed, and the battery is driving a current
counter-clockwise through the top loop. Suppose that the switch is opened at time t1,
help open until time t2, and then closed again. Describe the current that will be induced
in the bottom loop while all this occurs.
O There will be a brief pulse of counter-clockwise current, both when S is opened and when it is closed.
O There will be a brief pulse of counter-clockwise current when Sis opened, and then a brief pulse of clockwise current when S is closed.
O There will be a steady clockwise current before S is opened, a steady counter-clockwise current while S remains open, and then a steady clockwise current again after S is
closed.
O There will be a brief pulse of clockwise current, both when S is opened and when it is closed.
O There will be a brief pulse of clockwise current when S is opened, and then a brief pulse of counter-clockwise current when S is closed.
O There will be no current before S is opened, a steady counter-clockwise current while S is open, and then no current again after S is closed.
Transcribed Image Text:In the diagram at right, switch S is initially closed, and the battery is driving a current counter-clockwise through the top loop. Suppose that the switch is opened at time t1, help open until time t2, and then closed again. Describe the current that will be induced in the bottom loop while all this occurs. O There will be a brief pulse of counter-clockwise current, both when S is opened and when it is closed. O There will be a brief pulse of counter-clockwise current when Sis opened, and then a brief pulse of clockwise current when S is closed. O There will be a steady clockwise current before S is opened, a steady counter-clockwise current while S remains open, and then a steady clockwise current again after S is closed. O There will be a brief pulse of clockwise current, both when S is opened and when it is closed. O There will be a brief pulse of clockwise current when S is opened, and then a brief pulse of counter-clockwise current when S is closed. O There will be no current before S is opened, a steady counter-clockwise current while S is open, and then no current again after S is closed.
Expert Solution
Step 1

Initially, the switch was closed, so there was a uniform magnetic field because the current is constant. Thus, no changes in the magnetic field occur throughout the time until the switch is closed. After the switch is opened then there isn’t any current thus, nothing happens in this time span.

Step 2

At the time, when the switch is opened then the magnetic field decreases, and thus change in flux occurs. The changes in flux causes the induced current in the bottom loop. Using the Lenz law, the direction of the current in the bottom loop obtained as counter-clockwise.

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