Question

A circular coil having N turns is made from a wire L metre long. If a current of I ampere is passed through this coil suspended in a uniform magnetic field of B tesla, find the maximum torque that can act on this coil.

Answer 1

Hint: The product of the number of turns in the coil, the current, the magnetic field, the area of the coil, and the sine of the angle between the area vector and the magnetic field determines the torque produced in a coil placed in a magnetic field. We have all the necessary values to calculate the necessary counter torque value.

Complete step by step solution:
We know that a charged particle moving in a uniform magnetic field and the force acting on it is: the force exerted on a particle travelling with velocity in a uniform magnetic field and possessing charge q is defined by,
$\vec{F}=q(\vec{v}\times \vec{B})$
$\Rightarrow F=qvB\sin \theta $
Where θ is the angle between $\vec{v}$ and $\vec{B}$.

Case (i): If θ = 0, F = 0. Moreover, if = 180°, F = 0.
A charged particle experiences no force when it encounters a uniform magnetic field, whether it is moving in the same direction as the magnetic field or in the opposite direction.

Case (ii): F = qvB, if = 90°
In this instance, the particle is under the greatest possible force, which acts as a centripetal force to cause the charged particle to move in a circular motion.
$\therefore F=qvB=\dfrac{m{{v}^{2}}}{r}$
$\Rightarrow r=\dfrac{mv}{qB}$
where r denotes the path's circumference.

It is crucial to remember that this force is powerless to alter the charged particle's kinetic energy or speed. But because of the change in direction, it alters the charged particle's velocity, which also results in a change in momentum. Additionally, because the force is acting perpendicular to the direction of motion, it does no work. We know that torque is given by,
$\tau =\vec{M}\times \vec{B} \\ $
$\Rightarrow {{\tau }_{\max }}=MB \\ $
$\Rightarrow M=NIA=NI\pi {{R}^{2}} \\ $
Length = N × circumference
L = N2πR
$\Rightarrow R=\dfrac{L}{2\pi N} \\ $
$\Rightarrow M=NI\pi \dfrac{{{L}^{2}}}{4{{\pi }^{2}}{{N}^{2}}} \\ $
$\Rightarrow M=\dfrac{I{{L}^{2}}}{4\pi N} \\ $
$\therefore {{\tau }_{\max }}=\dfrac{I{{L}^{2}}B}{4\pi N}$

Hence, the maximum torque that can act on this coil is $\dfrac{I{{L}^{2}}B}{4\pi N}$.

Note: The force that can cause an object to rotate along an axis is measured as torque. In linear kinematics, force is what drives an object's acceleration. Similar to this, an angular acceleration is brought on by torque. As a result, torque can be thought of as the rotational counterpart to force.