Question

A bar magnet of length $14\, cm$ is placed in the magnetic meridian with its north pole pointing towards the geographic north pole. A neutral point is obtained at a distance of $18 cm$ from the center of the magnet. If $B _{ H }=0.4\, G$, the magnetic moment of the magnet is $\left(1\, G =10^{-4} T \right)$

• A. $2.880 \times 10^{3} J \, T ^{-1}$
• B. $2.880 \times 10^{2} J\, T ^{-1}$
• C. $2.880\, J\, T ^{-1}$
• D. $28.80\, J\, T ^{-1}$

Answer 1

Answer: (C)

i.e. $\frac{2 \mu_{0}}{4 \pi} \frac{ m }{ r ^{2}} \times \frac{7}{ r }=0.4 \times 10^{-4}$
$\Rightarrow 2 \times 10^{-7} \times \frac{ m \times 7}{\left(7^{2}+18^{2}\right)^{3 / 2}} \times 10^{4}$
$=0.4 \times 10^{-4}$
$m =\frac{4 \times 10^{-2} \times(373)^{3 / 2}}{14}$
$M = m \times 14 cm = m \times \frac{14}{100}$
$=\frac{0.04 \times(373)^{3 / 2}}{14} \times \frac{14}{100}$
$=4 \times 10^{-4} \times 7203.82=2.88 \,J / T$