Question

A freely suspended magnet oscillates with period T in earth’s horizontal magnetic field. When a bar magnet is brought near it, such that the magnetic field created by bar magnet is in same direction as earth’s horizontal magnetic field, the period decreases to T/2. The ratio of the field of the magnet F to the earth’s magnetic field (H) is(a) 1 : 3(b) 1 : 1(c) 3 : 1(d) 9 : 1

Answer 1

The ratio of the field of the magnet F to the earth’s magnetic field  is 3:1

Option (c) is correct.

Explanation:

We are given that:

• Time period = T
• Decrease in time period = T / 2

Solution:

Restoring torque due to the field T = −MB sinθ

For theta very small sinθ ≈ θ

T = −MBθ

T = Iα

α = − MBθ / I

Compare with α = −ω^2θ

ω = √MB / I

T = 1 / 2π √I / MB

T  = 1 / 2 π √ I / MB

T / 2 = 1 / 2π √I / M (B+B')

1 / 4π √I / MB = 1 / 2π √ I / M(B+B')

14.1 B = 1(B+B')

B + B' = 4 B

B' /B = 4 − 1 = 3

Thus the ratio of the field of the magnet F to the earth’s magnetic field  is 3:1