Torque and Work Done in a Magnetic Field

TORQUE ON A CURRENT CARRYING PLANAR LOOP IN A UNIFORM MAGNETIC FIELD

Case I.

When place of the loop is perpendicular to magnetic field

Length of AB = DC = $\ell$

And that of BC = AD = b

Forces experienced by all the sides are shown in the figure

$\therefore$ Force on AB and DC are equal and opposite to the each other and that on BC and AD too.

$\Rightarrow$ $\sum F=0$

Since the line of action of the forces on AB and DC is same and also the line of action of the forces BC and AD is same, therefore torque is zero.

Case II.

When the plane of the loop is inclined to the magnetic field.

In this case again $\sum F=0$

Lines of action of the forces on AB and DC are different, therefore this forms a couple and produces a torque. Side view of the loop is shown in the figure.

Torque = $Bi\ell (bsin\theta$

= $Bi(\ell b)sin\theta$

$\Rightarrow$ BiAsin $\theta$.

If loop has N turns then

$\tau$ = BNiA sin $\theta$

In vector form

$\vec{\tau }=\vec{M}\times \vec{B}$, where, $\vec{M}=NI\vec{A}$

Energy needed to rotate the loop through an angle d$\theta$ is

dU = $\tau$d$\theta$

$\Rightarrow$ $\Delta$U = $\int dU=\underset{{\theta }_1}{\overset{{\theta }_2}{\int }}\tau d\theta =\underset{{\theta }_1}{\overset{{\theta }_2}{\int }}MB\sin \theta d\theta$

$\Delta$U = MB(cos$\theta$₁ - cos$\theta$₂), if we choose at $\theta$₁ such that at $\theta$ = $\theta$₁, U₁ = 0

This is the energy stored in the loop.

U = -$\vec{M}.\vec{B}$

If a current carrying loop is placed in a uniform magnetic field it experiences a torque which is given by the expression,

$\overrightarrow{\tau }=\overrightarrow{M}x\overrightarrow{B}$

Illustration 1: A circular coil of wire 8 cm in diameter has 12 turns and carries a current of 5A. The coil is in a field where the magnetic induction is 0.6 T.

(i) What is the maximum torque on the coil?

(ii) In what position would the torque be one half as great as in (i)?

Solution: (i) The dipole moment of the coil = NiA

Here N = 12 turns, i = 5 amp, and

A = $\pi$r² = $\pi$(4 x 10 ^{– 2})²

$\therefore$ Dipole moment = 12 x 5 x $\pi$ x (4 x 10 ^{– 2})² = 0.30 Am²

Maximum torque = M_c B = 0.302 x 0.60 = 0. 181 n–m

(ii) If $\theta$ be the angle between the axis of the coil and the field

$\therefore$ torque = M_c B sin$\theta$

According to the problem, torque M_c B /2

$\therefore$ $\frac{M_{c}B}{2}=M_{c}B\sin \theta$

$\therefore$ $\sin \theta =\frac{1}{2}or\theta ={30}^{\circ }$

Thus the normal to the coil is at 30° to the field.

TORQUE ON A BAR MAGNET

Due to equal and opposite forces acting on the poles of a bar magnet placed in a uniform magnetic field B, a torque acts on it about an axis passing through the centre and perpendicular to the length given by

$\left|\vec{\tau }\right|=\left|\vec{M}\times \vec{B}\right|$=MB sin$\theta$

where $\theta$ is the angle between the magnetic field $\overrightarrow{B}$.and the magnetic moment $\overrightarrow{M}$.

Since the net force on the magnet is zero, the torque on the magnet causes oscillations about the equilibrium position

$\theta$ = 0º, when the magnet slightly displaced and set free.

If $I$ is the moment of inertia of the magnet and M its magnetic moment, for small angular displacement the time period of oscillations will be given by

T = 2$\pi$$\sqrt{\frac{I}{MB}}$ where moment of inertia $I$ = $\frac{m{\ell }^2}{12}$ for a thin magnet.

WORK DONE IN ROTATING A BAR MAGNET

The work done in rotating a bar magnet against the torque acting on it is

W = $\underset{{\theta }_1}{\overset{{\theta }_2}{\int }}\tau d\theta =\underset{{\theta }_1}{\overset{{\theta }_2}{\int }}MB\sin \theta d\theta$

= MB (cos$\theta$₁ - cos$\theta$₂)

The potential energy U = - $\vec{M}.\vec{B}$= - MB cos$\theta$

As work done is equal to change in potential energy

W = U₂ - U_­1 = MB(cos$\theta$₁ - cos$\theta$₂)

Illustration 2: A Magnet is suspended in the magnetic meridian with a untwisted wire. The upper end of the wire is rotated through 180° to deflect the magnet by 30° from magnetic meridian. Now this magnet is replaced by another magnet. Now the upper end of the wire is rotated through 270° to deflect the magnet by 30° from magnetic meridian. Compare the magnetic moments of magnets.

Solution: If f be the twist of the wire, then $\tau =C\varphi$, where C being restoring couple per unit twist of wire. Here

$\varphi$₁ = 180° – 30° = 150° = (150 x $\pi$/150) radian

$\varphi$₂ = (270° – 30° )= 240° = (240 x $\pi$/180) radian

If M be the magnetic moment and H, the horizontal component of earth's field, then $\tau$ = MH sin$\theta$

$\therefore C\varphi$ = MH sin$\theta$

If M₁ and M₂ be the magnetic moments of the two magnets respectively, then

$\therefore C\varphi$₁ = M₁H sin$\theta$

for first magnet

$\therefore C\varphi$₂ = M₂H sin$\theta$

for second magnet

$\therefore \frac{{\varphi }_1}{{\varphi }_2}=\frac{M_1}{M_2}or\frac{M_1}{M_2}=\frac{150\times \left(\pi /180\right)}{240\times \left(\pi /180\right)}=\frac{5}{8}]$

$\therefore$ M₁ : M₂ = 5 : 8