Question

A bar-magnet with magnetic moment 2.5 Am^2 is placed in a homogeneous magnetic field (of 0.1 T that is directed along the z-axis). What force and torque will the bar-magnet experience due to the magnetic field if: a) The north-pole of the magnet is pointing along the z-axis?

b) The north-pole of the magnet is pointing 45° relative to the z-axis?
c) The north-pole of the magnet is pointing 90° relative to the z-axis?

Answer 1

1)

Force on bar magnet = 0

Torque on bar magnet = 0

2)

Force on bar magnet = 0

Torque on bar magnet = 0.177 Nm

3)

Force on bar magnet = 0

Torque on bar magnet = 0.25 Nm

Step-by-step explanation:

Part 1)

net force on bar magnet in uniform magnetic field is always zero

Torque on bar magnet is given as

`\tau = MBsin\theta`

when bar magnet is inclined along z axis along magnetic field

then we will have

`\tau = MBsin0 = 0`

Part 2)

net force on bar magnet in uniform magnetic field is always zero

Torque on bar magnet is given as

`\tau = MBsin\theta`

when bar magnet is pointing 45 degree with z axis then we will have

`\tau = MBsin45`

`\tau = (2.5)(0.1)sin45`

`\tau = 0.177 Nm`

Part 3)

net force on bar magnet in uniform magnetic field is always zero

Torque on bar magnet is given as

`\tau = MBsin\theta`

when bar magnet is pointing 90 degree with z axis then we will have

`\tau = MBsin90`

`\tau = (2.5)(0.1)sin90`

`\tau = 0.25 Nm`