Question

The north pole of a bar magnet points towards a thin circular coil of wire containing 40 turns. The magnet is moved away from the coil, so that the ux through one turn inside the coil decreases by ∆Φ = 0.3 T.m2 in a time ∆t = 0.2 s. What is the average EMF induced in the (whole) coil during this time interval? Viewed from the side opposite of the bar magnet (from the right), does the induced current run clockwise or counterclockwise? Explain briey. [2 p.]

Answer 1

60 V

Current will flow anticlockwise from the right side of the magnet.

Explanation:

The number of turns on the coil = 40 turns

The magnetic flux changes by ∆Φ = 0.3 T.m^2

The time changes by ∆t = 0.2 s

The induced emf can be gotten as

E = N∆Φ/∆t

substituting values, we have

E = (40 x 0.3)/0.2 = 60 V

If we move the magnet away from the coil, then from Lenz law, the induced current on the coil will try to oppose the motion of the magnet by attracting the magnet towards the coil. For this to happen, the coil must possess the equivalent of a magnetic south pole. For the equivalent of a magnetic south pole, the current on the coil will flow in the clockwise direction when viewed from the left side of the magnet. This will appear as an anticlockwise direction, when viewed from the right side of the magnet.