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: Magnetic field due to a short magnetic dipole.
The magnetic field due to a short magnetic dipole is \[ \vec{B}=\frac{\mu_0}{4\pi d^3}\left[3(\vec{M}\cdot \hat{r})\hat{r}-\vec{M}\right] \] Let the two magnetic dipoles be along mutually perpendicular directions: \[ \vec{M}_1=M\hat{i} \] and \[ \vec{M}_2=M\hat{j} \]

Step 2: Direction of the point on the angle bisector.
Since the point lies on the bisector of the right angle, \[ \hat{r}=\frac{\hat{i}+\hat{j}}{\sqrt{2}} \]

Step 3: Field due to first dipole.
For \[ \vec{M}_1=M\hat{i} \] \[ \vec{M}_1\cdot \hat{r}=\frac{M}{\sqrt{2}} \] Therefore, \[ \vec{B}_1=\frac{\mu_0}{4\pi d^3} \left[ 3\left(\frac{M}{\sqrt{2}}\right) \left(\frac{\hat{i}+\hat{j}}{\sqrt{2}}\right)-M\hat{i} \right] \] \[ \vec{B}_1=\frac{\mu_0}{4\pi d^3} \left[ \frac{3M}{2}(\hat{i}+\hat{j})-M\hat{i} \right] \] \[ \vec{B}_1=\frac{\mu_0}{4\pi d^3} \left[ \frac{M}{2}\hat{i}+\frac{3M}{2}\hat{j} \right] \]

Step 4: Field due to second dipole.
For \[ \vec{M}_2=M\hat{j} \] Similarly, \[ \vec{B}_2=\frac{\mu_0}{4\pi d^3} \left[ \frac{3M}{2}\hat{i}+\frac{M}{2}\hat{j} \right] \]

Step 5: Resultant magnetic field.
\[ \vec{B}=\vec{B}_1+\vec{B}_2 \] \[ \vec{B}=\frac{\mu_0}{4\pi d^3} \left[ 2M\hat{i}+2M\hat{j} \right] \] Magnitude is \[ B=\frac{\mu_0}{4\pi d^3} \sqrt{(2M)^2+(2M)^2} \] \[ B=\frac{\mu_0}{4\pi d^3} \sqrt{8M^2} \] \[ B=\frac{\mu_0}{4\pi}\frac{2\sqrt{2}M}{d^3} \]

Step 6: Final conclusion.
Hence, the magnetic field is \[ \boxed{\frac{\mu_0}{4\pi}\frac{2\sqrt{2}M}{d^3}} \]