Q173 / 200 · Manipal (MET) 2014 · Physics
A short bar magnet is placed with its south pole towards geographical north. The neutral points are situated at a distance of 20 cm from the centre of the magnet. If $B_H=0.3 \times 10^{-4} \mathrm{~Wb} / \mathrm{m}^2$ then the magnetic moment of the magnet is
- A9000ab - amp $\times \mathrm{cm}^2$
- B$900 \mathrm{ab}-\mathrm{amp} \times \mathrm{cm}^2$
- Correct answerC1200ab - amp $\times \mathrm{cm}^2$
- D$225 a b-a m p \times \mathrm{cm}^2$
Solution
At neutral point magnetic field due to magnet = Horizontal component of the earth's magnetic field.
$\begin{aligned}
& \Rightarrow \frac{\mu_0}{4 \pi} \cdot \frac{2 M}{r^3}=B_H \Rightarrow \frac{10^{-7} \times 2 \times M \times 1}{(0.2)^3}=0.3 \times 15^4 \\
& \Rightarrow M=1.2 \mathrm{amp} \times \mathrm{m}^2=1200 \mathrm{ab}-\mathrm{amp} \times \mathrm{cm}^2
\end{aligned}$
Hence, the correct answer is option: 'C'.