Question

Two short magnets of equal dipole moments $M$ are fastened perpendicularly at their centres, as given in the figure. The magnitude of the magnetic field at a distance $d$ from the centre of the bisector of the right angle is -

[Point $\text{P}$ lies on the equatorial axis of both magnets]

• A. $\frac{{\mu }_{0}M}{4\pi d^3}$
• B. $\frac{\sqrt{2}{\mu }_{0}M}{4\pi d^3}$
• C. $\frac{2\sqrt{2}{\mu }_{0}M}{4\pi d^3}$
• D. $\frac{2{\mu }_{0}M}{4\pi d^3}$

Answer 1

The correct option is B $\frac{\sqrt{2}{\mu }_{0}M}{4\pi d^3}$

Field due to a short bar magnet at a point on its equatorial axis is,

$B=\frac{{\mu }_{0}M}{4\pi d^3}$


Since, the given point $\text{P}$ is on the equatorial axis of the given both bar magnets.

The magnetic field at $\text{P}$ due to the horizontal magnet is,

${\overrightarrow{B}}_1=\frac{{\mu }_{0}M}{4\pi d^3}\hat{i}$

The magnetic field at $\text{P}$ due to the vertical magnet is,

${\overrightarrow{B}}_2=\frac{{\mu }_{0}M}{4\pi d^3}(-\hat{j})$

Hence, net magnetic field at point $\text{P}$ will be

${\overrightarrow{B}}_{net}={\overrightarrow{B}}_1+{\overrightarrow{B}}_2$

$|{\overrightarrow{B}}_{net}|=\sqrt{B_1^2+B_2^2}=\frac{\sqrt{2}{\mu }_{0}M}{4\pi d^3}$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(B\right)$ is the correct answer.