A stiff copper wire is balanced horizontally on a pivot, as shown in Fig. 8.1.

Question 13

(a) Define magnetic flux density. [3]

(b) A stiff copper wire is balanced horizontally on a pivot, as shown in Fig. 8.1.

Fig. 8.1

Sections PQ, QR and RS of the wire are situated in a uniform magnetic field of flux density B produced between the poles of a permanent magnet.

The perpendicular distance of PQRS from the pivot is 7.5 cm.

When a current of 2.7 A is passed through the wire, a small mass of 45 mg is placed a distance 8.8 cm from the pivot in order to restore the balance of the wire, as shown in Fig. 8.2.

Fig. 8.2

(i) Explain why, when the current is switched on, the current in the sections PQ and RS of the wire does not affect the balance of the wire. [2]

(ii) The length of section QR of the wire is 1.2 cm.

Calculate the magnetic flux density B. [3]

[Total: 8]

Reference: Past Exam Paper – November 2018 Paper 42 Q8

Solution:

(a) Magnetic flux density is the force per current per unit length of a wire (conductor) when the current (in the wire / conductor) is at right angles to the magnetic field.

(b)

(i)

The forces (on PQ and RS) are horizontal.

Hence they create no moment about the pivot.

{The wire is balance when the clockwise moment on it is equal to the anticlockwise moment.

Moment = force × perpendicular distance from line of action of force to pivot

In this case, the forces are horizontal – there is not PERPENDICULAR distance. So, the forces do not create any moment.}

(ii)

need to apply moments

{From the principle of moments,

Moment due to the force on QR about the pivot = Moment due to the weight of the mass of 45 mg

(BIL) × x = (mg) × y               

where x = 7.5 cm (perpendicular distance of the force BIL from the pivot)

y = 8.8 cm (perpendicular distance of the weight from the pivot)}

            L = 1.2 cm = 1.2×10^‑2 m

            m = 45 mg = 45×10^-6 kg

BILx = mgy

B × 2.7 × 1.2×10^‑2 × 7.5 = 45×10^-6 × 9.81 × 8.8

B = 1.6 × 10^-2 T