Two coils P and Q are placed close to one another, as shown in Fig. 10.1.

Question 3

Two coils P and Q are placed close to one another, as shown in Fig. 10.1.

Fig. 10.1

(a) The current in coil P is constant.

An iron rod is inserted into coil P.

Explain why, during the time that the rod is moving, there is a reading on the voltmeter

connected to coil Q. [2]

(b) The current in coil P is now varied as shown in Fig. 10.2.

Fig. 10.2

On Fig. 10.3, show the variation with time of the reading of the voltmeter connected to coil Q for time t = 0 to time t = t2. [4]

Fig. 10.3

[Total: 6]

Reference: Past Exam Paper – June 2016 Paper 42 Q10

Solution:

(a)

The iron rod changes the flux (density)/field as it moves.

{From Faraday’s law,} a change of the flux linkage in coil Q causes an induced e.m.f.

(b)

- constant reading (either polarity) from time zero to near t1

{A change in current causes the flux to change. This induces an e.m.f. From time zero to near t₁, the increase in current is constant (straight line, positive gradient), so the e.m.f. induced is almost constant.}

- spike in one direction near t1 clearly showing a larger voltage

{Around t₁, there is a sharp decrease in current. So, the e.m.f. induced is of a greater magnitude (greater than the previous one) as the change in current is very large. This can be shown by a spike.}

- of opposite polarity  

{From time zero to neat t₁, the current was increasing. So, the e.m.f. induced is in one direction (let’s say positive). Around t₁, the current decreases. So, the e.m.f. induced is now of the opposite polarity (negative).}

- zero reading from near t1 to t2

{From t₁ to t₂, the current is constant – that is, it is NOT changing. So, there is no change in flux. No e.m.f. is induced.}