Question

A bar magnet of dipole moment 3 Amº rests with its centre on a
frictionless pivot. A force Fis 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How will the equilibrium be effected if Fis withdrawn?​

Answer 1

Answer:

External force applied on magnet is 3.75 N an if the force is withdrawn then magnet will not remain in equilibrium

Explanation:

As we know that magnetic bar is in equilibrium under external force

So we will have

$\tau_{mag} = \tau_{force}$

so we have

$MBsin\theta = FL$

now plug in all values

$3(0.25)sin30 = F(0.10)$

$0.375 = F(0.10)$

$F = 3.75 N$

If force F is withdrawn then the magnetic torque is not balanced and hence the magnet is not remain in equilibrium

#Learn

Topic : Magnetic torque

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