Question

Two short magnets of equal dipole moments \(M\) are fastened perpendicularly at their centers. The magnitude of the magnetic field at a distance \(d\) from the centre on the bisector of the right angle is \((\mu_0=\text{Permeability of free space})\)

• A. \(\dfrac{\mu_0}{4\pi}\dfrac{2\sqrt{2}M}{d^3}\)
• B. \(\dfrac{\mu_0}{4\pi}\dfrac{5M}{d^3}\)
• C. \(\dfrac{\mu_0}{4\pi}\dfrac{2M}{d^3}\)
• D. \(\dfrac{\mu_0}{4\pi}\dfrac{10M}{d^3}\)

Hint:

For a magnetic dipole, use \[ \vec{B}=\frac{\mu_0}{4\pi r^3}\left[3(\vec{M}\cdot \hat{r})\hat{r}-\vec{M}\right] \] and add the magnetic fields vectorially when more than one dipole is present.

Answer 1

The Correct Option is A

Step 1: Set up the coordinate system.
Let the two magnetic dipoles be oriented along mutually perpendicular directions: $\vec{M}_1 = M\hat{i}$ and $\vec{M}_2 = M\hat{j}$. The point $P$ on the bisector of the right angle between them is along the direction $\hat{r} = \dfrac{\hat{i} + \hat{j}}{\sqrt{2}}$ at distance $d$ from the center.
Step 2: Recall the general formula for the magnetic field of a short dipole.
The magnetic field at position $\vec{r}$ due to a magnetic dipole $\vec{M}$ is: \[ \vec{B} = \frac{\mu_0}{4\pi r^3}\left[3(\vec{M}\cdot\hat{r})\hat{r} - \vec{M}\right] \]
Step 3: Calculate the field due to dipole $\vec{M}_1$.
$\vec{M}_1 \cdot \hat{r} = M \cdot \dfrac{1}{\sqrt{2}} = \dfrac{M}{\sqrt{2}}$. So: \[ \vec{B}_1 = \frac{\mu_0}{4\pi d^3}\left[3 \cdot \frac{M}{\sqrt{2}} \cdot \frac{\hat{i}+\hat{j}}{\sqrt{2}} - M\hat{i}\right] = \frac{\mu_0 M}{4\pi d^3}\left[\frac{3}{2}(\hat{i}+\hat{j}) - \hat{i}\right] = \frac{\mu_0 M}{4\pi d^3}\left[\frac{1}{2}\hat{i} + \frac{3}{2}\hat{j}\right] \]
Step 4: Calculate the field due to dipole $\vec{M}_2$.
By symmetry ($M_2$ is along $\hat{j}$, $M_2 \cdot \hat{r} = \dfrac{M}{\sqrt{2}}$): \[ \vec{B}_2 = \frac{\mu_0 M}{4\pi d^3}\left[\frac{3}{2}\hat{i} + \frac{1}{2}\hat{j}\right] \]
Step 5: Add the two field vectors.
\[ \vec{B} = \vec{B}_1 + \vec{B}_2 = \frac{\mu_0 M}{4\pi d^3}\left[\left(\frac{1}{2}+\frac{3}{2}\right)\hat{i} + \left(\frac{3}{2}+\frac{1}{2}\right)\hat{j}\right] = \frac{\mu_0 M}{4\pi d^3}\left[2\hat{i} + 2\hat{j}\right] \] \[ |\vec{B}| = \frac{\mu_0 M}{4\pi d^3} \times 2\sqrt{2} \]
Step 6: State the final answer.
The magnitude of the resultant magnetic field at the bisector point is: \[ \boxed{|\vec{B}| = \frac{\mu_0}{4\pi}\frac{2\sqrt{2}\,M}{d^3}} \]