A short bar magnet of magnetic moment $5.25 × 10^{-2}\, J/T$ is placed with its axis perpendicular to earth's magnetic field direction. The distance of the point from the center of magnet on its perpendicular bisector, where the resultant field is 45° to the earth's field will be: (Earth's field at that place is 0.42 G)

$5 × 10^{-2} m$

The correct answer is Option (2) → $5 × 10^{-2} m$

Magnetic field due to a short bar magnet on its perpendicular bisector at distance $r$:

$B_m = \frac{\mu_0}{4\pi} \frac{M}{r^3}$

Given: $M = 5.25 \times 10^{-2}\ \text{J/T}$, Earth's field $B_e = 0.42\ \text{G} = 0.42 \times 10^{-4}\ \text{T}$

Resultant field makes angle $45^\circ$ with Earth's field:

$\tan 45^\circ = \frac{B_m}{B_e} = 1 \text{ implies } B_m = B_e$

So, $\frac{\mu_0}{4\pi} \frac{M}{r^3} = B_e \quad$ (no extra factor for perpendicular bisector)

$r^3 = \frac{\mu_0}{4\pi} \frac{M}{B_e}$

Substitute values: $\frac{\mu_0}{4\pi} = 10^{-7}$

$r^3 = \frac{10^{-7} \cdot 5.25 \times 10^{-2}}{0.42 \times 10^{-4}} = \frac{5.25 \times 10^{-9}}{0.42 \times 10^{-4}} \approx 1.25 \times 10^{-4}$

$r = (1.25 \times 10^{-4})^{1/3} \approx 0.05\ \text{m} = 5\ \text{cm}$

Final Answer: $r \approx 5\ \text{cm}$