Brendan

Brendan 4 Jun 2017 - 2

Question

A copper bar of length L is moving to the right with a uniform speed v in a region of uniform magnetic field of flux density B, directly perpendicularly downwards into the paper in the figure below.

The ends of the rods are rigidly connected to a voltmeter which moves with the rod. What is the reading on the voltmeter?

A) zero
B) non-zero reading less than Blv
C) BLv
D) more than BLv


Answer

Method 1:

Consider wires PQ and RS. Since both wires are moving through the magnetic field, both of them will have an induced e.m.f. of BLv.  

Based on Fleming’s Right Hand rule, the induced current of PQ and RS will both be upwards.

This means that the resultant current will be zero and the e.m.f. measured by the voltmeter will be zero.

Answer is A.


Method 2:

Consider the closed loop PQRS. As the loop moves through the magnetic field, the magnetic flux linkage remains the same since the area of the loop does not change.

Since there is no change in the magnetic flux linkage, by Faraday’s law there is no induced e.m.f. and the voltmeter will read 0 V.

Answer is A.

Brendan 4 Jun 2017 - 1

Question

An e.m.f. is induced in a coil placed in a changing magnetic field. The flux density B of this field varies with time as shown below.

Brendan-20170604-01.jpg

At which value of t is the magnitude of the e.m.f. induced in the wire a maximum?

A) 1 ms
B) 2 ms
C) 3 ms
D) 4 ms


Answer

The induced e.m.f. is given by the formula:

Hence the magnitude of the induced e.m.f. is given by:

This means that the induced e.m.f. will be the greatest when the rate of change of the magnetic flux density is the greatest.

From the graph, the rate of change of the magnetic flux density is the greatest when the gradient of the graph is the greatest.

This means that the induced e.m.f. will be the greatest at time t = 0 ms and t = 4 ms.

Hence the answer is D.