A horseshoe magnet is placed on a balance. A stiff metal wire is clamped horizontally between the poles, as illustrated in Fig. 5.1.

Question 6

(a) Define the tesla. [3]

(b) A horseshoe magnet is placed on a balance. A stiff metal wire is clamped horizontally between the poles, as illustrated in Fig. 5.1.

Fig. 5.1

The magnetic flux density in the space between the poles of the magnet is uniform and

is zero outside this region.

The length of the metal wire normal to the magnetic field is 6.4 cm.

When a current in the wire is switched on, the reading on the balance increases by 2.4 g.

The current in the wire is 5.6 A.

(i) State and explain the direction of the force on the wire due to the current. [3]

(ii) Calculate the magnitude of the magnetic flux density between the poles of the

magnet. [2]

(c) A low frequency alternating current is now passed through the wire in (b).

The root-mean-square (r.m.s.) value of the current is 5.6 A.

Describe quantitatively the variation of the reading seen on the balance. [2]

Reference: Past Exam Paper – June 2012 Paper 42 Q5

Solution:

(a) The tesla is defined as the uniform magnetic flux density of a magnetic field to which a (long) straight (conductor) wire, carrying a current of 1 A, is normal, causing a force per unit length of 1 N m^-1 (for a flux density of 1 T).

(b)

(i)

{We want the direction of the force on the wire.

Since the reading on the balance increases, we can conclude that}

there is originally a downward force on the horseshoe magnet (due to the current).

{This is the force on the horseshoe magnet. But we want the force on the wire.

From Newton’s third law, there are 2 forces, equal in magnitude and opposite in direction, acting on different bodies. Since the force on the magnet is downwards, the force on the wire (a different body) is opposite – that is upwards.}

From Newton’s third law, there will be an upward force on the wire.

(ii)

F = BIL

{The force is due to the increase in weight. This can be obtained from the increase in mass from the balance reading.

Force F = weight = mg = 2.4×10^-3 × 9.8 }

2.4×10^-3 × 9.8 = B × 5.6 × 6.4×10^-2   

Magnetic flux density, B = 0.066 T (need 2 s.f.)

(c)

{The maximum current I₀ = √2 × I_rms }

The new reading on the balance is now 2.4√2 = 3.4 g.

{Since an alternating current is now used, the direction of the force on the wire / magnet alternates up and down such that at some instant, the balance reading increases by 3.4 g and at some instant, the balance reading decreases by 3.4 g.

The total change in the reading = 3.4 – -3.4 = 6.8 g}

EITHER The reading changes between +3.4 g and -3.4 g

OR The total change is 6.8 g