Torque experienced by a Magnet in a Magnetic Field and Tangent Law

Torque experienced by a Magnet in a Magnetic Field

Suppose a bar magnet suspended freely in a uniform magnetic field B. let m be the strength of each pole and 2l its length. The force acting on N-pole is mB along the field and that in S-pole mB opposite to the field. These two forces form a couple and rotate the magnet so as to align it along the field B. The torque of a couple acting on the bar magnet is

$$\begin{align*} \tau &= mBl\: \sin\: \theta + mBl\sin \: \theta \\ &= 2mlB\sin \: \theta \\ &= MB\: \sin \: \theta \\ \end{align*}$$

where $M = m\times 2l$, the magnetic moment of the magnet and $\theta $, the angle initially made by the magnet with B. fundamentally, the torque arises from the interaction between B and the magnetic field of the bar magnet. In vector form,

$$\begin{align*} \vec {\tau }&= \vec M \times \vec B \\ \text {The torque is maximum when} \: \theta \:\text {is} \: 90^o \: \text {and so} \\ \tau &= M\: B.\\ \text {When} \: B = 1 \: \text {unit, then} \\ \tau &= M \\ \end{align*}$$

So, the magnetic moment of a dipole is numerically equal to the torque acting on the dipole when it is placed perpendicular to the direction of uniform magnetic field of unit strength.

In this reference, the unit of magnetic moment is joule per tesla,$JT^{-1} )$.

The work done by the torque in turning the magnet through an angle $\theta $ is

$$\begin{align*} \Delta W &= \tau \Delta \theta \\ &= MB \: \sin\theta \: Delta \theta \\ \\ \text {and total work done on turning from} \: \theta_1 \: \text {to} \: \theta _2 \\ W &= \int \Delta W \\ &= \int_{\theta _1}^{\theta_2} MB \sin \theta \: d\theta \\ &= - MB (\cos \: \theta _2 - \cos \: \theta_1 ) \\ \end{align*}$$

The work done by external torque is stored in the magnet as potential energy, U. If the potential energy at $90^0 $ is zero, then at $ \theta _2 = \theta $ is

$$ U = - MB \: \cos \: \theta = -\vec M. \vec B.$$

The change in potential energy when the magnet is rotated through $180^o $ is

$$U = MB – (-MB) = 2MB $$

Tangent Law

When a bar magnet is freely suspended in a region where two horizontal, uniform magnetic fields are perpendicular to each other, the bar magnet comes in equilibrium under the action of two couples field and remains at rest in a definite direction between the fields.

Suppose two uniform horizontal magnetic fields B and H perpendicular to each other as shown in the figure. Here 2l be the length and m the pole strength of the bar magnet which is freely suspended in couple acting on the magnet due to the B is $mB \times SA $ and tends to rotate the magnet in anticlockwise direction. While the torque due to the field H is $mH \times NA $ and tends to rotate the magnet in clockwise direction. At equilibrium position, two torques are equal and opposite and then

$$\begin{align*} m\: B \times SA &=mH \times NA \\ \text {or,} \: B &= H \frac {SA}{NA} \\ \frac {SA}{NA} &= \tan \: \theta \: \text {and} \\ B &= H\tan \: \theta \\ \end{align*}$$

This is the tangent law and is used in deflection magnetometer.