The magnetic field of a short bar magnet at a point $10$ cm from its centre on the axial line is $2 \times 10^{-4}$ T. The field at a point $20$ cm from its centre on the equatorial line is
Answer: (B) $1.25 \times 10^{-5}$ T
Concept. For a short magnet, $B_{\text{axial}} = \dfrac{\mu_0}{4\pi}\dfrac{2M}{r^3}$ and $B_{\text{eq}} = \dfrac{\mu_0}{4\pi}\dfrac{M}{r^3}$.
Step 1. Find $\dfrac{\mu_0 M}{4\pi}$. At $r = 0.1$ m on the axis:
$$\frac{\mu_0}{4\pi}\frac{2M}{(0.1)^3} = 2 \times 10^{-4} \Rightarrow \frac{\mu_0 M}{4\pi} = 10^{-7}\ \text{T m}^3$$
Step 2. Equatorial field at $0.2$ m.
$$B = \frac{10^{-7}}{(0.2)^3} = \frac{10^{-7}}{8 \times 10^{-3}} = 1.25 \times 10^{-5}\ \text{T}$$
Answer. $1.25 \times 10^{-5}$ T (Option B).
Solution by Sreeraj P, M.Sc Physics