Question

A circular loop of area \[1c{m^2}\], carrying a current of $10\,A$, is placed in a magnetic field of $0.1\,T$ perpendicular to the plane of the loop. The torque on the loop due to the magnetic field is?
A. Zero
B. \[{10^{ - 4}}N - m\]
C. \[{10^{ - 2}}N - m\]
D. 1 N m

Answer 1

Hint:Torque on the current loop, also known as the magnetic moment, is the product of the current in the loop, the area of cross-section made by the loop, the number of loops that are present and the magnetic field on the loop.

Formula Used:
\[T = NBiA\sin \theta \]
where $T$ is the torque, $N$ is the number of loops, $B$ is the magnetic field, $i$ is the current, A is the area, and \[\theta \] is the angle.

Complete step by step solution:
We have been given that the area of the loop is \[A = 1\,c{m^2}\], carrying a current $i = 10\, A$, placed in a magnetic field of $B = 0.1\,T$ and we have to find the torque on loop due to magnetic field.

We can say that the number of loops is $N = 1$, and as there is a whole loop, we can say that the angle made by the loop is 360 degrees. We can find the torque on the loop as,
\[T = NBiA\sin \theta \\
\Rightarrow T = 1 \times 0.1 \times 10 \times 1 \times \sin \left( {{{360}^ \circ }} \right) \\
\Rightarrow T = 1 \times 0 \\
\therefore T = 0 \]

So, option A zero is the required torque on the loop or required solution.

Note: The angle of loop made should always be mentioned while finding the torque of loop due to its magnetic field. Always see that the measurements are in the same unit to make any calculations.