$A$ circular coil carrying a current of $2.5 \ A$ is free to rotate about an axis in its plane perpendicular to an external magnetic field. When the coil is made to oscillate, the time period of oscillation is $T$. If the current through the coil is $10 \ A$, the time period of oscillation is:

$A$ current loop in a magnetic field:

$(a)$ $A$ circular coil of $30$ turns and radius $8.0 \; cm$ carrying a current of $6.0 \; A$ is suspended vertically in a uniform horizontal magnetic field of magnitude $1.0 \; T$. The field lines make an angle of $60^{\circ}$ with the normal of the coil. Calculate the magnitude of the counter torque that must be applied to prevent the coil from turning.
$(b)$ Would your answer change,if the circular coil in $(a)$ were replaced by a planar coil of some irregular shape that encloses the same area? (All other particulars are also unaltered.)

$A$ current-carrying loop is placed in a uniform magnetic field. The torque acting on it does not depend upon

$A$ thin non-conducting disc of radius $R$ is rotating clockwise (see figure) with an angular velocity $\omega$ about its central axis,which is perpendicular to its plane. Both its surfaces carry positive charges of uniform surface density. Half the disc is in a region of a uniform,unidirectional magnetic field $B$ parallel to the plane of the disc,as shown. Then,

$A$ square loop of side $2a$ carrying current $I$ is kept in the $XZ$ plane with its centre at the origin. $A$ long wire carrying the same current $I$ is placed parallel to the $z$-axis and passes through the point $(0, b, 0)$,where $b >> a$. The magnitude of the torque on the loop about the $z$-axis is given by: