When the switch in the drawing is closed, the current in the coil increases to its equilibrium value. While the current is increasing there is an induced current in the metal ring. The ring is free to move. What happens to the ring? a. It does not move b. It jumps upward c. It jumps downward

Question

Giovanna Crane

Physics

19 January, 20:50

When the switch in the drawing is closed, the current in the coil increases to its equilibrium value. While the current is increasing there is an induced current in the metal ring. The ring is free to move. What happens to the ring? a. It does not move b. It jumps upward c. It jumps downward

+4

Answers (1)

Know the Answer?

Rocco Ruiz · Answer 1

the previous correct answer is b

Explanation:

When the circuit is closed in the system, a current is induced that follows the lenz law, which opposes the change that is occurring and therefore the coil increases and the idicidal current in the ring must reach the maximum oppositing is the current of the coil, so quiet force is repulsion

Consequently, the previous correct answer is b