91Ó°ÊÓ

1. Figure (a) with a rectangular loop having forces acting on all the sides of the rectangle.
2. Figure (b) with right hand thumb rule acting on the rectangular loop.
3. Figure (c) with a rod placed in a uniform magnetic field.

A magnetic field exerts a force on a straight wire carrying current; it exerts a torque on a loop of wire carrying a current that results due to the inducement of the magnetic force about the radial length of the conductor. Torque causes an object to spin around a fixed axis due to its action by the applied force along the radial vector of the conductor. We use the equation of torque applied by the magnetic field on a current carrying loop. Upon replacing the loop of arbitrary shape with several adjacent long, thin, approximately rectangular loops that we can assume that the distribution of the current is not altered. Then we can prove the relation using the integration method.

Formula:

The torque acting at a point inside a magnetic field,

$\mathrm{Ï„}=NiABsin\mathrm{θ}$ …(¾±)

where, N is the number of turns in the coil, i is the current of the wire, A is the area of the conductor, B is the magnetic field,$\mathrm{θ}$is the angle made by the conductor with the magnetic field.

Let us replace the current loop of arbitrary shape with an assembly of thin, adjacent, and small rectangular loops such as to cover the same area enclosed by the original loop. We can assume that each rectangular loop carries the same current ‘i’ as that flowing through the original loop. So, the magnitude of the small torque exerted by the magnetic field B on the n^th rectangular loop will be given using equation (i) as follows:

$∆\stackrel{â‡\P Ä}{\mathrm{Ï„}}=NiB\sin \mathrm{θ}∆A_n$

So, to get the torque due to a whole assembly of rectangular loops we need to do summation of all torques. Thus, integrating the above equation will result in the value of the net torque of the arbitrary loop as follows:

role="math" localid="1662889959685" $$\begin{gathered} \mathrm{τ}=NiABsin\mathrm{θ}×\underset{n}{∑}∆A_n \\ \mathrm{τ}=NiB\sin \mathrm{θ}×A \\ \mathrm{τ}=NiAB\sin \mathrm{θ} \end{gathered}$$

Hence, the value of the torque is, $\mathrm{τ}=NiAB\sin \mathrm{θ}$.