$A$ 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:

  • A
    $\frac{2 \mu_{0} I^{2} a^{2}}{\pi b}$
  • B
    $\frac{\mu_{0} I^{2} a^{3}}{2 \pi b^{2}}$
  • C
    $\frac{\mu_{0} I^{2} a^{2}}{2 \pi b}$
  • D
    $\frac{2 \mu_{0} I^{2} a^{3}}{\pi b^{2}}$

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The figure below shows the north and south poles of a permanent magnet in which an $n$-turn coil 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 $B$. The torque on the coil is:

The square loop $ABCD$,carrying a current $i$,is placed in a uniform magnetic field $B$,as shown. The loop can rotate about the axis $XX'$. The plane of the loop makes an angle $\theta$ $(\theta < 90^o)$ with the direction of $B$. Through what angle will the loop rotate by itself before the torque on it becomes zero?

$A$ 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 $60^o$ with the normal of the coil. Calculate the magnitude of the counter torque that must be applied to prevent the coil from turning. (in $, Nm$)

The radius of a circular ring of wire is $R$ and it carries a current of $I \, A$. At its centre,a smaller ring of radius $r$ with current $i$ and $N$ turns is placed. Assuming that the planes of the two rings are perpendicular to each other and the magnetic induction produced at the centre of the bigger ring is constant,then the torque acting on the smaller ring will be:

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