#### Question

A magnetised needle of magnetic moment 4.8 × 10^{−2} JT^{−1} is placed at 30° with the direction of uniform magnetic field of magnitude 3 × 10^{−2} T. Calculate the torque acting on the needle.

#### Solution

The torque (τ) of a needle in uniform magnetic field will be given as

`vectau = vecM xx vecB =MB sintheta`

Here

M = magnetic moment = 4.8 ×10^{−2} J/T

B = magnetic field strength = 3 × 10^{−2} T

θ = angle with respect to field = 30°

So, the torque will be rewritten as

`vectau`= (4.8 ×10^{−2}) × (3 × 10^{−2}) sin 30°

`or vectau = 14.4 xx 10^-4 xx 1/2 = 7.2 xx 10^-4 N`

Is there an error in this question or solution?

Solution A Magnetised Needle of Magnetic Moment 4.8 × 10−2 Jt−1 is Placed at 30° with the Direction of Uniform Magnetic Field of Magnitude 3 × 10−2 T. Calculate the Torque Acting on the Needle. Concept: Torque on Current Loop, Magnetic Dipole - Torque on a Rectangular Current Loop in a Uniform Magnetic Field.