$A$ circular coil of radius $R$ and a current $I$,which can rotate about a fixed axis passing through its diameter,is initially placed such that its plane lies along the magnetic field $B$. The kinetic energy of the loop when it rotates through an angle $90^{\circ}$ is: (Assume that $I$ remains constant)

$A$ small circular loop of conducting wire has radius $a$ and carries current $I$. It is placed in a uniform magnetic field $B$ perpendicular to its plane such that when rotated slightly about its diameter and released,it starts performing simple harmonic motion of time period $T$. If the mass of the loop is $m$,then find the expression for $T$.

$A$ solenoid of length $0.4 \ m$ and having $500$ turns of wire carries a current $3 \ A$. $A$ thin coil having $10$ turns of wire and radius $0.1 \ m$ carries current $0.4 \ A$. The torque required to hold the coil in the middle of the solenoid with its axis perpendicular to the axis of the solenoid is $(\mu_0 = 4\pi \times 10^{-7} \ SI \ units, \pi^2 = 10, \sin 90^{\circ} = 1)$.

$A$ closely wound solenoid of $2000$ turns and area of cross-section $1.5 \times 10^{-4} \, m^2$ carries a current of $2.0 \, A$. It is suspended through its centre and perpendicular to its length,allowing it to turn in a horizontal plane in a uniform magnetic field of $5 \times 10^{-2} \, T$ making an angle of $30^o$ with the axis of the solenoid. The torque on the solenoid will be:

$A$ coil carrying electric current is placed in a uniform magnetic field,then

$A$ square coil of side $10 \ cm$ having $200$ turns is placed in a uniform magnetic field of $2 \ T$ such that the plane of the coil is in the direction of the magnetic field. If the current through the coil is $3 \ mA$, then the torque acting on the coil is