# Torque on a current carrying loop in a uniform magnetic field

(i) When a coil carrying current is placed in uniform magnetic field, the net force on it is zero but it experiences a torque or couple. (ii)Torque action on a current carrying coil placed in uniform magnetic field is $\large \vec{\tau} = \vec{M} \times \vec{B}$

(iii)Torque acting on the coil is t =  N I A B cosα  = N I A B sinθ ; Here

A = area of coil carrying current I

N = number of turns of the coil

B = Magnetic induction of the field

α = Angle made by the plane of the coil with $\vec{B}$

θ = Angle made by the normal to the plane of the coil with

(iv)If the plane of coil is parallel to the direction of magnetic field ,  τmax = NIAB

(v)If the plane of coil is perpendicular to the direction of magnetic field, τ = 0

(vi)If current carrying coil is placed in a non-uniform magnetic field it experiences both force and torque.

(vii)For a given area, torque is independent of shape of the coil

(viii)Torque is directly proportional to area of the coil.

Energy needed to rotate the loop through an angle dθ is

$\displaystyle dU = \vec{\tau} .\vec{d\theta}$

$\displaystyle dU = \int dU = \int_{\theta_1}^{\theta_2} \tau d\theta$

$\displaystyle U = \int_{\theta_1}^{\theta_2} M B sin\theta d\theta$

$\displaystyle U = -M B (cos\theta_2 – cos\theta_1)$

This is the energy stored in the loop.

Special cases:

(i) When a current carrying coil is placed in uniform magnetic field, net force on it F = 0. But net torque may acts.

(ii) When a current carrying coil is placed in nonuniform magnetic field, net force, net torque both acts.

tnet ≠ 0         ;           Fnet ≠ 0

(iii) If the angel made by $\vec{M}$ of the coil with  $\vec{B}$ in uniform magnetic field is ‘θ’, then its potential energy $\large P.E = – \vec{M}.\vec{B} = = -MBcos\theta$

(iv)If a current carrying coil is rotated in a uniform field such that the angle made by $\vec{M}$ with $\vec{B}$ is changes from θ1 to θ2. W = -MB (cosθ2 – cosθ1)

(v)If ext field is along, the direction of $\vec{M}$ , then θ = 0° , τ = 0 , P.E = -MB (min)

This position corresponds to stable equilibrium.

(vi)If external magnetic field is opposite to then, θ = 180°

τ = 0     ;   P.E. = + MB (max)

This corresponds to unstable equilibrium.

Illustration : A uniformly charged disc whose total charge has magnitude q and whose radius is r rotates with constant angular velocity of magnitude ω. What is the magnetic dipole moment ? Solution: The surface charge density is q/πr2. Hence the charge within a ring of radius R and width dR is

$\displaystyle dq = \frac{q}{\pi r^2}(2 \pi R dR)$

$\displaystyle dq = \frac{2 q}{r^2}(R dR)$

The current carried by this ring is its charge divided by the rotation period,

$\displaystyle di = \frac{dq}{2\pi/\omega} = \frac{q\omega}{\pi r^2} (R dR)$

The magnetic moment contributed by this ring has the magnitude dM = A |di|, where a is the area of the ring.

$\displaystyle dM = \pi R^2 |di|= \frac{q\omega}{ r^2} (R^3 dR)$

$\displaystyle M = \int dM = \int_{0}^{r} \frac{q\omega}{ r^2} (R^3 dR)$

$\displaystyle M = \frac{q\omega r^2}{4}$

Exercise : Find the maximum torque on a rectangular coil of 85 turns each of dimensions 0.2 m by 0.3 m, carrying a current of 2.0 A in a uniform magnetic field B = 6.5 T