Question

A bar magnet of length 6 cm is placed in the magnetic meridian with N pole, pointing towards the geographical north. Two neutral points, separated by a distance of 8 cm are obtained on the equatorial axis of the magnet. If \( B_H = 1.2 \times 10^{-5} \, \text{T} \), then the pole strength of the magnet is

• A. 0.75 A-m\(^2\)
• B. 0.25 A-m\(^2\)
• C. 0.50 A-m\(^2\)
• D. 1.50 A-m\(^2\)

Hint:

Use the formula for the magnetic field on the equatorial axis of a dipole to find the pole strength. The distance between the neutral points helps determine the value.

Answer 1

The Correct Option is B

The magnetic field strength \( B_H \) at a neutral point on the equatorial axis of a dipole is given by: \[ B_H = \frac{\mu_0 m}{4\pi r^3} \] where \( m \) is the pole strength, \( r \) is the distance from the center of the magnet, and \( \mu_0 \) is the permeability of free space. Given the distance between the neutral points as 8 cm, and using the provided value of \( B_H \), we can solve for the pole strength \( m \). The result comes out to be \( 0.25 \, \text{A-m}^2 \). Therefore, the correct answer is (B).