$A$ $250$-turn rectangular coil of length $2.1\, cm$ and width $1.25\, cm$ carries a current of $85\, \mu A$ and is subjected to a magnetic field of strength $0.85\, T.$ The work done to rotate the coil by $180^o$ against the torque is .............. $\mu J$.

$A$ circular coil of $20$ turns and radius $10\; cm$ is placed in a uniform magnetic field of $0.10\; T$ normal to the plane of the coil. If the current in the coil is $5.0\; A$,what is the
$(a)$ total torque on the coil,
$(b)$ total force on the coil,
$(c)$ average force on each electron in the coil due to the magnetic field?
(The coil is made of copper wire of cross-sectional area $10^{-5} \;m ^{2},$ and the free electron density in copper is given to be about $10^{29}\; m ^{-3}.$)

$A$ circular coil having $N$ turns and radius $R$ carrying current $I$ is held in the $z-x$ plane in a magnetic field $B\hat{k}$. The torque on the coil due to the magnetic field in $N-m$ is:

$A$ current-carrying loop is free to turn in a uniform magnetic field. The loop will come into equilibrium when its plane is inclined at

$A$ coil having $100$ turns, area of $5 \times 10^{-3} \, m^2$, carrying current of $1 \, mA$ is placed in a uniform magnetic field of $0.20 \, T$ such that the plane of the coil is perpendicular to the magnetic field. The work done in turning the coil through $90^{\circ}$ is . . . . . . $\mu J$.

The given figure shows the north and south poles of a permanent magnet in which a coil of $n$ turns of cross-sectional area $A$ is placed. When a current $I$ is passed through the coil,the plane of the coil makes an angle $\theta$ with respect to the direction of the magnetic field $B$. If the plane of the magnetic field and the coil are horizontal and vertical respectively,the torque on the coil will be