$STATEMENT-1$ $A$ vertical iron rod has a coil of wire wound over it at the bottom end. An alternating current flows in the coil. The rod goes through a conducting ring as shown in the figure. The ring can float at a certain height above the coil. Because
$STATEMENT-2$ In the above situation,a current is induced in the ring which interacts with the radial component of the magnetic field to produce an average force in the upward direction.

  • A
    Statement-$1$ is True,Statement-$2$ is True; Statement-$2$ is a correct explanation for Statement-$1$.
  • B
    Statement-$1$ is True,Statement-$2$ is True; Statement-$2$ is $NOT$ a correct explanation for Statement-$1$.
  • C
    Statement-$1$ is True,Statement-$2$ is False.
  • D
    Statement-$1$ is False,Statement-$2$ is True.

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Two different coils have inductances $L_1 = 4 \ mH$ and $L_2 = 2 \ mH$. At a certain instant,the current in the two coils is increasing at the same constant rate,and the power supplied to the first coil is four times that of the second coil. If $e$,$I$,and $U$ indicate the potential difference,current,and energy stored in the inductance respectively,then which of the following is correct?

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$A$ metallic square wire loop of side $10\, cm$ and resistance $1\,\Omega$ is moved with a constant velocity $v_0$ in a uniform magnetic field of induction $B = 2\, T$ as shown in the figure. The magnetic field is perpendicular into the plane of the loop. The loop is connected to a network of resistors each of value $3\,\Omega$. The resistance of the lead wires $OS$ and $PQ$ are negligible. What should be the speed of the loop so as to have a steady current of $1\, mA$ in it? Give the direction of current in the loop.

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The current $i$ in an inductance coil varies with time $t$ according to the following graph. Which one of the following plots shows the variations of voltage $V$ in the coil?

$A$ rectangular loop of sides $12 \text{ cm}$ and $5 \text{ cm}$,with its sides parallel to the $x$-axis and $y$-axis respectively,moves with a velocity of $5 \text{ cm/s}$ in the positive $x$-axis direction in a space containing a variable magnetic field in the positive $z$-direction. The field has a gradient of $10^{-3} \text{ T/cm}$ along the negative $x$-direction and it is decreasing with time at the rate of $10^{-5} \text{ T/s}$. If the resistance of the loop is $6 \text{ m}\Omega$,the power dissipated by the loop as heat is . . . . . . $\times 10^{-9} \text{ W}$.

$A$ 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 center at the origin. $A$ magnetic dipole of moment $m$ is brought along the axis of this loop from infinity to a point at distance $r \gg a$ from the center of the loop with its north pole always facing the loop,as shown in the figure.
The magnitude of the 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
$(A) \frac{m}{r^3} \quad (B) \frac{m^2}{r^2} \quad (C) \frac{m}{r^2} \quad (D) \frac{m^2}{r}$
$(2)$ 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
$(A) \frac{m}{r^5} \quad (B) \frac{m^2}{r^5} \quad (C) \frac{m^2}{r^6} \quad (D) \frac{m^2}{r^7}$

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