A special metal S conducts electricity without any resistance. A closed wire loop, made of S, does not allow any change in flux through itself by inducing a suitable current to generate a compensating flux. The induced current in the loop cannot decay due to its zero resistance. This current gives rise to a magnetic moment which in turn repels the source of magnetic field or flux. Consider such a loop, of radius a, with its centre at the origin. A magnetic dipole of moment $m$ is brought along the axis of this loop from infinity to a point at distance $r(>>$ a) from the centre of the loop with its north pole always facing the loop, as shown in the figure below.
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 $\frac{ km _{1} m _{2}}{ r ^{4}}$, where $k$ is a constant of appropriate dimensions. The direction of this force is along the line joining the two dipoles.
$\frac{m }{ r ^{3}}$
\(\frac{m^2}{r^2}\)
\(\frac{m}{r^2}\)
\(\frac{m^2}{r}\)
The correct answer is (A) :\(\frac{m }{ r ^{3}}\)
\(\frac{m}{r^5}\)
\(\frac{m^2}{r^5}\)
\(\frac{m^2}{r^6}\)
\(\frac{m^2}{r^7}\)
The correct answer is (C): \(\frac{m^2}{r^6}\)
Two identical concave mirrors each of focal length $ f $ are facing each other as shown. A glass slab of thickness $ t $ and refractive index $ n_0 $ is placed equidistant from both mirrors on the principal axis. A monochromatic point source $ S $ is placed at the center of the slab. For the image to be formed on $ S $ itself, which of the following distances between the two mirrors is/are correct:
The major products obtained from the reactions in List-II are the reactants for the named reactions mentioned in List-I. Match each entry in List-I with the appropriate entry in List-II and choose the correct option.