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): \(\frac{m^2}{r^6}\)
Let $ S $ denote the locus of the point of intersection of the pair of lines $$ 4x - 3y = 12\alpha,\quad 4\alpha x + 3\alpha y = 12, $$ where $ \alpha $ varies over the set of non-zero real numbers. Let $ T $ be the tangent to $ S $ passing through the points $ (p, 0) $ and $ (0, q) $, $ q > 0 $, and parallel to the line $ 4x - \frac{3}{\sqrt{2}} y = 0 $.
Then the value of $ pq $ is
Let $ y(x) $ be the solution of the differential equation $$ x^2 \frac{dy}{dx} + xy = x^2 + y^2, \quad x > \frac{1}{e}, $$ satisfying $ y(1) = 0 $. Then the value of $ 2 \cdot \frac{(y(e))^2}{y(e^2)} $ is ________.