
$B_{1}=B_{2}$
At point $P$
The magnetic field will be in same direction by both coils,
$B=B_{1}+B_{2}$
$=B_{1}+B_{1}$
$=2 B_{1}$
The magnetic field at point $P$ is,
$B=2\left(\frac{\mu_{0} i r^{2}}{2\left(r^{2}+n^{2}\right)^{3 / 2}}\right)$
$=\frac{\mu_{0} i r^{2}}{2\left(r^{2}+n^{2}\right)^{3 / 2}}$
Substitute the values in the above equation.
$B=\frac{\mu_{0} \times \frac{7}{2} \times 0.1^{2}}{2\left(0.1^{2}+0.05^{2}\right)^{3 / 2}}$
$=\frac{56 \mu_{0}}{\sqrt{5}} T$


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 ($2$)
(permeability of free space $\left.=4 \pi \times 10^{-7} H / m \right)$