Question

A circular loop of radius R carries a current I. Another circular loop of radius r(< < R) carries a current i and is placed at the centre of the larger loop. The planes of the two circles are at right angles to each other. Find the torque acting on the smaller loop.

Answer 1

Hint:Use the formula for the magnitude of the torque applied on a current carrying loop when it is placed in an external uniform magnetic field. Here, the magnetic field will be the magnetic field produced by the larger loop at its centre.

Formula used:
$B=\dfrac{{{\mu }_{0}}I}{2R}$
$\Rightarrow\tau =iAB\sin \theta $
where $i$ is the current in the loop, $A$ is the area enclosed by the loop and $B$ is the magnetic field and $\theta $ is the angle between the area vector and the magnetic field.

Complete step by step answer:
The larger loop of radius R will produce a magnetic field at its centre. It is given that the current in the loop is I. Then the magnetic field at its centre will be equal to,
$B=\dfrac{{{\mu }_{0}}I}{2R}$ …. (i).
The direction of the magnetic field will be perpendicular to the plane of the loop. The direction is given by the right hand thumb rule.Since the outside loop is much larger than the loop at its centre, the magnetic field through the smaller loop can be considered as constant.When a loop is placed in a uniform magnetic field, the magnetic field exerts a torque on the loop. The magnitude of the torque is given as,
$\tau =iAB\sin \theta $ ….. (ii),
where i is the current in the loop, A is the area enclosed by the loop and B is the magnetic field and $\theta $ is the angle between the area vector and the magnetic field.
In this case, $A=\pi {{r}^{2}}$, $B=\dfrac{{{\mu }_{0}}I}{2R}$.
It is given that the planes of the both the loops are perpendicular. Since the magnetic field is perpendicular to a larger loop and the area vector is perpendicular to the smaller, the area vector is perpendicular to the magnetic field.
This means that $\theta ={{90}^{\circ }}$.
Substitute the values in (ii).
$\tau =i\left( \pi {{r}^{2}} \right)\left( \dfrac{{{\mu }_{0}}I}{2R} \right)\sin {{90}^{\circ }}$
$\therefore \tau =\dfrac{{{\mu }_{0}}\pi iI{{r}^{2}}}{2R}$.

Therefore, the torque on the loop is equal to $\dfrac{{{\mu }_{0}}\pi iI{{r}^{2}}}{2R}$.

Note:Torque is the measure of the force that can cause an object to rotate about an axis. Force is what causes an object to accelerate in linear kinematics. Similarly, torque is what causes an angular acceleration. Hence, torque can be defined as the rotational equivalent of linear force.The torque on the loop placed in an external magnetic field in vector form is given as $\tau =i\left( \overrightarrow{A}\times \overrightarrow{B} \right)$.
A current carrying loop acts as a magnetic dipole. The magnetic moment of the loop is given as $\overrightarrow{M}=i\overrightarrow{A}$
Therefore, the torque can be written as $\tau =\overrightarrow{M}\times \overrightarrow{B}$.