A long solenoid has an area of cross-section of 28 cm2, as shown in Fig. 5.1.

Question 7

(a) Define the tesla. [2]

(b) A long solenoid has an area of cross-section of 28 cm², as shown in Fig. 5.1.

Fig. 5.1

A coil C consisting of 160 turns of insulated wire is wound tightly around the centre of

the solenoid.

The magnetic flux density B at the centre of the solenoid is given by the expression

B = μ0n I

where I is the current in the solenoid, n is a constant equal to 1.5 × 10³ m^-1 and μ0 is

the permeability of free space.

Calculate, for a current of 3.5 A in the solenoid,

(i) the magnetic flux density at the centre of the solenoid, [2]

(ii) the flux linkage in the coil C. [2]

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

(ii) The current in the solenoid in (b) is reversed in direction in a time of 0.80 s.

Calculate the average e.m.f. induced in coil C. [2]

Reference: Past Exam Paper – June 2013 Paper 41 & 43 Q5

Solution:

(a) The tesla is the (uniform magnetic) flux is normal to a long (straight) wire carrying a current of 1 A, creates a force per unit length of 1 N m^–1.

(b)

(i)

{Magnetic flux density B = μ₀n I}

Magnetic flux density B = (4π×10^–7) × (1.5×10³) × 3.5

Magnetic flux density B = 6.6 × 10^–3 T

(ii)

{Flux linkage = Magnetic Flux × Number of turns

Flux linkage = BA × N

Cross-sectional area A should be in SI units.}

Flux linkage = (6.6 × 10^–3) × (28 × 10^–4) × 160

Flux linkage = 3.0 × 10^–3 Wb

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

(ii)

{To reverse the direction of the current, the current must first be reduced to zero, and then increased to the same value in the opposite direction. So, the flux linkage calculated above should be  considered twice.

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

Average e.m.f. = (2 × 3.0 × 10^–3) / 0.80

Average e.m.f. = 7.4 × 10^–3 V