
In given case,
$d=R \sin 45^{\circ}=\frac{R}{\sqrt{2}}$
$\theta_{1}=135^{\circ}, \theta_{2}=180^{\circ}$
$\therefore B($ at $P)=\frac{\mu_{0} I}{4 \pi\left(\frac{R}{\sqrt{2}}\right)}\left[\cos 135^{\circ}-\cos 180^{\circ}\right]$
$=\frac{\mu_{0} I}{4 \pi R} \sqrt{2}\left(\frac{-1}{\sqrt{2}}-(-1)\right)$
$=\frac{\mu_{0} I}{4 \pi R} \sqrt{2}\left(\frac{\sqrt{2}-1}{\sqrt{2}}\right)$
or $B($ at $P)=\frac{\mu_{\rho} I}{4 \pi R}(\sqrt{2}-1) T$
The magnitude of magnetic field of a dipole $m$, at a point on its axis at distance $r$, is $\frac{\mu_0}{2 \pi} \frac{m}{r^3}$, where $\mu_0$ is the permeability of free space. The magnitude of the force between two magnetic dipoles with moments, $m_1$ and $m_2$, separated by a distance $r$ on the common axis, with their north poles facing each other, is $\frac{k m_1 m_2}{r^4}$, where $k$ is a constant of appropriate dimensions. The direction of this force is along the line joining the two dipoles.
($1$) When the dipole $m$ is placed at a distance $r$ from the center of the loop (as shown in the figure), the current induced in the loop will be proportional to
$(A)$ $\frac{m}{r^3}$ $(B)$ $\frac{m^2}{r^2}$ $(C)$ $\frac{m}{r^2}$ $(D)$ $\frac{m^2}{r}$
($2$) The work done in bringing the dipole from infinity to a distance $r$ from the center of the loop by the given process is proportional to
$(A)$ $\frac{m}{r^5}$ $(B)$ $\frac{m^2}{r^5}$ $(C)$ $\frac{m^2}{r^6}$ $(D)$ $\frac{m^2}{r^7}$
Give the answer or qution ($1$) and ($2$)


