A magnetised needle of magnetic moment 4.8 x 10–2 J T–1 is placed

1.	A magnetised needle of magnetic moment 4.8 x 10–2 J T–1 is placed at 30° with the direction of uniform magnetic field of magnitude 3 x 10–2 T. Calculate the torque acting on the needle.
Answer» We have, τ = MB sin θ Where τ → torque acting on magnetic needle M → Magnetic moment B → Magnetic field strength Then τ = 4.8 x 10^–2 x 3 x 10^–2 sin 30° = 4.8 x 10^–2 x 3 x 10^–2 x \(\frac{1}{2}\) = 7.2 x 10^–4Nm

A magnetised needle of magnetic moment 4.8 x 10–2 J T–1 is placed at 30° with the direction of uniform magnetic field of magnitude 3 x 10–2 T. Calculate the torque acting on the needle.

Answer»

We have, τ = MB sin θ

Where τ → torque acting on magnetic needle

M → Magnetic moment

B → Magnetic field strength

Then τ = 4.8 x 10^–2 x 3 x 10^–2 sin 30°

= 4.8 x 10^–2 x 3 x 10^–2 x \(\frac{1}{2}\)

= 7.2 x 10^–4Nm