A bar magnet of length 14 cm is placed in the magnetic meridian with its north pole pointing geographic north. A neutral point is found at 18 cm from the centre. Given B_H = 0.4 text G (1 text G = 10^ -4 text T ), find the magnetic moment of the magnet.

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EEJ MAIN Physics QUESTION #7012

Question 1

A bar magnet of length 14 cm is placed in the magnetic meridian with its north pole pointing geographic north. A neutral point is found at 18 cm from the centre. Given $B_H = 0.4\ \text{G}$ ($1\ \text{G} = 10^{-4}\ \text{T}$), find the magnetic moment of the magnet.

$2.880 \times 10^2\ \text{J\,T}^{-1}$
$2.880\ \text{J\,T}^{-1}$
$2.880 \times 10^3\ \text{J\,T}^{-1}$
$28.80\ \text{J\,T}^{-1}$✔️

Correct Answer Explanation

N-pole points north: neutral points lie on the equatorial line (broadside position) where equatorial field equals $B_H$.

With $d = 18\ \text{cm} = 0.18\ \text{m}$, $l = 7\ \text{cm} = 0.07\ \text{m}$, $B_H = 0.4\times10^{-4}\ \text{T}$:

Equatorial field: $B = \frac{\mu_0}{4\pi}\frac{M}{(d^2+l^2)^{3/2}}$

$M = \frac{B_H(d^2+l^2)^{3/2}}{10^{-7}} = \frac{0.4\times10^{-4}\times(0.0324+0.0049)^{3/2}}{10^{-7}} \approx 28.80\ \text{J\,T}^{-1}$

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