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Two short magnets of equal dipole moments $M$ are fastened perpendicularly at their centres. The magnitude of the magnetic field at a distance $d$ from the centre on the bisector of the right angle is ( $\mu_0=$ Permeability of free space)
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The correct answer is:
$\frac{\mu_0}{4 \pi} \frac{2 \sqrt{2} M}{d^3}$
When two short magnets of equal dipole moments $M$ are fastened perpendicularly at their centres. Then their resultant dipole moment is given as
$\begin{aligned} M^{\prime} & =\sqrt{M_1^2+M_2^2+2 M_1 M_2 \cos \theta} \\ & =\sqrt{M^2+M^2+2 M M \cos 90^{\circ}}\end{aligned}$
$\Rightarrow M^{\prime}=M \sqrt{2}$ ...(i)
The magnitude of the magnetic field at a distance $d$ from the centre on the bisector i.e at a distance $d$ on the axial position of dipole,
$B=\frac{\mu_0}{4 \pi} \cdot \frac{2 M^{\prime}}{d^3}$
$=\frac{\mu_0}{4 \pi} \cdot \frac{2 \sqrt{2} M}{d^3}$ [from Eq. (i)]
$\begin{aligned} M^{\prime} & =\sqrt{M_1^2+M_2^2+2 M_1 M_2 \cos \theta} \\ & =\sqrt{M^2+M^2+2 M M \cos 90^{\circ}}\end{aligned}$
$\Rightarrow M^{\prime}=M \sqrt{2}$ ...(i)
The magnitude of the magnetic field at a distance $d$ from the centre on the bisector i.e at a distance $d$ on the axial position of dipole,
$B=\frac{\mu_0}{4 \pi} \cdot \frac{2 M^{\prime}}{d^3}$
$=\frac{\mu_0}{4 \pi} \cdot \frac{2 \sqrt{2} M}{d^3}$ [from Eq. (i)]
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