Two short magnets of equal dipole moments $M$ are fastened perpendicularly at their centres which lies at origin. Let two magnets lie along $X$-axis and $Y$-axis, respectively. The magnitude of the magnetic field at a distance $R$ from the centre on the $Y$-axis is $\frac{\mu_0}{4 \pi} \frac{M_0}{R^3}$. Assuming, $R>>l$ (magnet length), the magnitude of $M$ is

• A

  $\frac{M_0}{2 \sqrt{2}}$
• B

  $\frac{M_0}{2}$
• Correct answer
  C

  $\frac{M_0}{\sqrt{5}}$
• D

  $\frac{M_0}{\sqrt{2}}$

Solution

Two magnet are joined as shown in figure. The magnetic field at point $P$ due to magnet $M_1$ is on axial line, while that due to $M_2$ is on equatorial line.
So, net magnetic field at a point $P$ at a distance $R$ on $Y$-axis, $B_{\text {net }}=\sqrt{B_1^2+B_2^2}=\sqrt{\left(\frac{\mu_0}{4 \pi} \frac{2 M}{R^3}\right)^2+\left(\frac{\mu_0}{4 \pi} \frac{M}{R^3}\right)^2}$ $B_{\text {net }}=\frac{\mu_0}{4 \pi R^3} \cdot \sqrt{5} M$$\ldots$ (i) Given, $B_{\text {net }}=\frac{\mu_0}{4 \pi R^3} \cdot M_0$ $\ldots$ (ii) By comparision Eq. (i) and Eq. (ii), we get $M=\frac{M_0}{\sqrt{5}}$

Hence, the correct answer is option: 'C'.