Question

Find the formula for the couple acting on a magnetic dipole placed in a magnetic field. On its basis, define magnetic dipole moment.

Hint:

Treat the bar magnet as two poles \(\pm q_m\); the equal and opposite pole forces form a couple of moment \(mB\sin\theta\), so \(\vec{\tau}=\vec{m}\times\vec{B}\).

Answer 1

Step 1: Set up the dipole in the field.
Consider a magnetic dipole (a small bar magnet) of magnetic length \(2l\) and pole strength \(q_m\). Its magnetic dipole moment is \(m = q_m \times 2l\). Place it in a uniform magnetic field \(\vec{B}\) so that its axis makes an angle \(\theta\) with the field.

Step 2: Find the forces on the two poles.
The force on a pole of strength \(q_m\) in field \(B\) is \(F = q_m B\).
The north pole feels a force \(q_m B\) along \(\vec{B}\); the south pole feels a force \(q_m B\) opposite to \(\vec{B}\). These forces are equal, opposite and act along different lines, so their resultant force is zero but they form a couple.

Step 3: Moment (torque) of the couple.
Torque = (either force) \(\times\) (perpendicular distance between the two forces). The perpendicular distance between the lines of action is \(2l\sin\theta\).
\[ \tau = q_m B \times 2l \sin\theta \] Step 4: Introduce the dipole moment.
Since \(m = q_m \times 2l\), \[ \tau = m B \sin\theta \] In vector form \(\vec{\tau} = \vec{m} \times \vec{B}\).

Step 5: Define magnetic dipole moment.
Putting \(\theta = 90^\circ\) and \(B = 1\) gives \(\tau = m\). So the magnetic dipole moment is the torque acting on the dipole when it is held perpendicular to a magnetic field of unit strength. \[\boxed{\vec{\tau} = \vec{m}\times\vec{B},\quad \tau = mB\sin\theta}\]