Question

A bar magnet of dipole moment 3 Am2 rests with its centre on a
frictionless pivot. A force F is applied at right angles to the axis of
the magnet, 10 cm from the pivot. It is observed that an external
magnetic field of 0.25 T is required to hold the magnet in equilibrium
at an angle of 30° with the field.
Calculate the value of F.
How will the equilibrium be effected if F is withdrawn ?
13
Р.​

Answer 1

Given :

Dipole moment of given bar magnet resting on the frictionless pivot = $3Am^2$

Distance from the pivot where a force F is applied at right angles to the axis = 10 cm

The magnitude of external magnitude field required to hold the magnet in equilibrium at an angle of 30° = 0.25 T

To Find ;

Value of F = ?

What is the effect on equilibrium when F is withdrawn ?

Solution :

We know that torque on a magnetic dipole of magnetic moment 'm' in an uniform magnetic field B is given as :

$\vec\tau = \vec m \times \vec B$    -(1)

If $\theta$ is the angle between $\vec m$ and $\vecB$$\vec B$ , we can write (1) as :

$\vec\tau = \vec m .\vec B . sin\theta$

On putting the values we get :

$\vec\tau = 3 \times 0.25 \times sin30$

$\vec\tau= \frac{3}{8} Nm$        -(2)

Now the torque due to the applied force :

$\vec\tau'= \vec r \times \vec F = |\vec r ||\vec F| sin \theta'$

Here $\theta'$ is 90°

$|\vec\tau|' = 10 \times 10 ^{-2} \times F= \frac{F}{10}$     -(3)

Since for equilibrium $\tau _{net} = 0$ , so :

$|\tau| = |\tau|'$

Or ,$\frac{3}{8} = \frac{F}{10}$

Or , F= $\frac{15}{4} N$

So,  value of F is $\frac{15}{4} N$ .

And if the F is withdrawn , the dipole will rotate in the direction of applied  torque.