Question

A circular coil having N turns and radius r carries a current I. It is held in the XZ plane in a magnetic field \[B\widehat{i}\]. The torque on the coil due to magnetic field is:
A. \[B\pi {{r}^{2}}IN\]
B. \[\dfrac{B{{r}^{2}}I}{\pi N}\]
C. zero
D. \[\dfrac{B\pi {{r}^{2}}I}{N}\]

Answer 1

Hint: In this question we have been asked to calculate the torque on the given coil due to the magnetic field. We have been given the magnetic field of the coil. We know that torque is a vector quantity. Therefore, it has direction as well as magnitude. Since we have been asked to calculate only the magnitude, we shall use the formula for torque on a coil. Which deals with magnetic moments and the magnetic field.

Formula Used:
 \[\tau =MB\sin \theta \]
Where,
\[\tau \] is the torque on the coil.
M is the magnetic moment
B is the magnetic field
\[M=NIA\]

Complete step by step answer:
We have been asked to calculate the magnitude of the torque on the coil due to the given magnetic field.

Now, from the formula of torque we know,
\[\tau =MB\sin \theta \]
Now, we have been asked to calculate only the magnitude
Therefore,
We can say that
\[\left| \tau \right|=M\times B\] ………….. (1)
Solving for M
We know that for a current carrying coil the magnetic moment of a coil is given by,
\[M=NIA\]
We know that
\[A=\pi {{r}^{2}}\]
Therefore,
\[M=NI\pi {{r}^{2}}\] ………….. (2)
Substituting (2) in equation (1)
We get,
\[\tau =NI\pi {{r}^{2}}\times B\]
Therefore,
\[M=B\pi {{r}^{2}}IN\]
Therefore, the correct answer is option A.

Note:
In a current carrying coil a magnetic field is generated. This magnetic field produces a torque on the coil, which is given as the product of the number of turns of coil, the current in the coil, area of the loop and the magnetic field. To calculate the direction of the torque the, we can calculate the cross product of The magnetic moment vector and the magnetic field vector.