Q 12-05-011NEETJEE MainMedium
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
$B_{axial} = \dfrac{2kM}{r^3}$ and $B_{eq} = \dfrac{kM}{r^3}$. At twice the distance, $r^3$ is $8$ times, and equatorial is half of axial:
$$B = \frac{2 \times 10^{-4}}{2 \times 8} = 1.25 \times 10^{-5}\ \text{T}$$
Solution by Sreeraj P, M.Sc Physics