Two EAMCET 2004 Questions on Magnetism

Even though isolated magnetic monopoles are not available, you will find questions involving isolated north poles and south poles. The practical method of obtaining a pole of the required polarity at a point is to use a long bar magnet so that the effect of the other pole at the point can be neglected. Consider the following MCQ which appeared in EAMCET 2004 question paper

Two magnetic isolated north poles each of strength ‘m’ ampere-metre are placed one at each of two vertices of an equilateral triangle of side ‘a’. the resultant magnetic induction at the third vertex is

(a) μ₀m/4π a² (b) μ₀√2 m/4π a² (c) μ₀√3 m/4π a² (d) μ₀m/4π a

The magnetic inductions (flux density) at the third vertex due to the two isolated poles have the same magnitude B = μ₀m/4π a², but their directions are inclined at 60º. The magnitude of the resultant flux density at the third vertex is

[B² + B² + 2B² cos 60º]^1/2 = √3 B = μ₀√3 m/4π a²

[If the triangle is right angled and isosceles with short sides of length ‘a’ and the poles are placed at the ends of the long side, the flux density at the third vertex will be μ₀√2m/4πa²]

The following MCQ also appeared in EAMCET 2004 question paper:

A bar magnet used in a vibration magnetometer is heated so, as to reduce the magnetic moment by 36%. The time period of magnet (neglecting the changes in dimensions of magnet)

(a) increases by 25 % (b) decreases by 36 %

(c) increases by 16 % (d) decreases by 16 %

The period (T) of angular oscillations of the magnet in a magnetic field of flux density ‘B’ is given by

T = 2π√(I/mB) where ‘I’ is the moment of inertia of the magnet about the suspension fibre and ‘m’ is the magnetic dipole moment.

When the dipole moment is reduced by 36%, ‘m’ is to be replaced by 0.64m in the above equation. The period of oscillation then becomes T/(0.8) = 1.25T.

The period thus increases by 25%.