A coil C having 120 turns of wire is wound on to one end of the core. The area of cross-section of coil C is 1.5 cm2.

Question 8

(a) State Faraday’s law of electromagnetic induction. [2]

(b) A solenoid S is wound on a soft-iron core, as shown in Fig. 9.1.

Fig. 9.1

A coil C having 120 turns of wire is wound on to one end of the core. The area of cross-section of coil C is 1.5 cm².

A Hall probe is close to the other end of the core.

When there is a constant current in solenoid S, the flux density in the core is 0.19 T. The

reading on the voltmeter connected to the Hall probe is 0.20 V.

The current in solenoid S is now reversed in a time of 0.13 s at a constant rate.

(i) Calculate the reading on the voltmeter connected to coil C during the time that the current is changing. [2]

(ii) Complete Fig. 9.2 for the voltmeter readings for the times before, during and after the

direction of the current is reversed.

Fig. 9.2

[4]

 [Total: 8]

Reference: Past Exam Paper – November 2018 Paper 41 & 43 Q9

Solution:

(a) Faraday’s law of electromagnetic induction states that the (induced) e.m.f. is proportional to the rate of change of (magnetic) flux linkage.

(b)

(i)

{induced e.m.f. = change in flux linkage / time

Change in flux linkage = BAN}

induced e.m.f. = (ΔB)AN / Δt

{As the current is reversed, the flux density B is also reversed. That is, it changes from 0.19 T to {to zero and from zero to} – 0.19 T. The change in flux density = 2 × 0.19 T}

induced e.m.f. = (2 × 0.19 × 1.5×10^-4 × 120) / 0.13

induced e.m.f. = 0.053 V                                           

(ii)

reading on voltmeter connected to coil C / V:            0          0.053               0

{For an e.m.f. to be induced, there should be a change in flux density. So, before the current changes and after the current changes, there is no change in current and thus, no change in flux density. So, no e.m.f. is induced.

Note that there IS a current in these cases, but the current is NOT changing.

During the change in current, the induced e.m.f. is 0.053 V (as calculated above).}

reading on voltmeter connected to Hall probe / V: zero in middle column

final column correct sign (negative)                                                              

final column correct magnitude (0.20)                                                          

{Hall voltage = BI / ntq

When the current I is zero, the Hall voltage is also zero. (middle column).

When the current is reversed, its direction changes. So, the Hall voltage will be in the reversed direction – a negative sign should be included.

Hall voltage after current changes = - 0.20 V}