c
(c) Let us suppose a bar magnet of length \(2l\) is placed inside the magnetic field \((B).\)

Torque\((t)\) \(=\) Force acting \(\times\) perpendicular distance \((d).\)

Force acting \(=B m\)

\(\sin \theta=\frac{d}{2 l}\)

\(T=B m \times d\)

\(d=2 l \sin \theta\)

\(T=B m \cdot 2 l \sin \theta\)

\(M=m \times 2 l T=M B \sin \theta\)

\(T=-\vec{B} \times \vec{M}\)

\(T=\vec{M} \times \vec{B}\)

Where \(; B=\) Magnetic field of induction.

\(M=\) magnetic moment.

\(m=\) pole strength of bar magnet.

\(d=\) perpendicular distance.

\(2 l=\) length of bar magnet.

\(T=\) torque.

Hence, the answer is \(\vec{M} \times \vec{B}\).