MAGNETIC DIPOLE (BAR MAGNET) IN A UNIFORM MAGNETIC FIELD :
Derive an expression for torque acting on a magnetic dipole placed in magnetic field.
Consider a magnetic dipole (a bar magnet) placed in a uniform magnetic field \( \vec{B} \) such that the angle between the direction of magnetic dipole moment \( (\vec{m}) \) and the direction of magnetic field \( (\vec{B}) \) is \( \theta \)
Let:
- Magnetic Length of magnet = \( 2l \)
- Pole Strength of each pole = \( q_m \)
- Force acting on North pole = \( q_m B \) along the direction of \( \vec{B} \)
- Force acting on South pole = \( q_m B \) opposite to the direction of \( \vec{B} \)
These two equal and opposite forces constitute a couple which tends to rotate the magnet in the direction of \( \vec{B} \). Thus, the bar magnet experiences a torque and tends to rotate. However, net force acting on the magnetic dipole is zero and hence magnetic dipole does not have the translational motion in the magnetic field.
Torque acting on the bar magnet or magnetic dipole is given by:
$\tau$ = Magnitude of Force × Perpendicular distance between forces
$\tau= q_m B \times ZN$
$\because \text{in } \triangle SZN, \sin \theta = \frac{ZN}{SN}$
$ZN = SN \sin \theta$
$\tau= qm B (\text{SN} \sin \theta)$
$\tau= q_m B (2l \sin \theta)$
$\tau= (q_m \times 2l) B \sin \theta$
$(\because q_m \times 2l = m)$
or
$\tau = m.B \sin \theta \tag{1}$
In vector form:
$\vec{\tau} = \vec{m} \times \vec{B} \tag{2}$
Direction of torque:
The direction of torque is perpendicular to the plane containing \(\vec{m}\) and \(\vec{B}\)
Definition of magnetic dipole moment:
We know,
$\tau = mB \sin \theta$
When \(B = 1\) unit and \(\theta = 90^\circ\), then
$\tau = m \quad (\because \sin 90^\circ = 1)$
Thus, magnetic dipole moment can be defined as the torque acting on a magnetic dipole placed normal to a uniform magnetic field of unit strength.
Unit of magnetic dipole moment:
$m = \frac{\tau}{B \sin \theta} = \frac{\text{Nm}}{\text{T}} = \text{NmT}^{-1} \quad \text{or} \quad JT^{-1} \quad \text{or} \quad Am^2$