A magnetic needle suspended parallel to a magnetic field requires √ 3 J of work to turn it through 60°. The torque needed to maintain the needle in this position will be

Question

A magnetic needle suspended parallel to a magnetic field requires √ 3 J of work to turn it through 60°. The torque needed to maintain the needle in this position will be (A) 2 √3 J (B) √3 J (C) 3 J (D) (3/2) J

Tardigrade Admin · Accepted Answer

Work done in changing the orientation of a magnetic needle of magnetic moment

M

in a magnetic field

B

from position

θ_{1}

to

θ_{2}

is given by

W = MB (cos θ_{1} - cos θ_{2})

Here,

θ_{1} = 0^{\circ}, θ_{2} = 6 0^{\circ}

= MB (1 - \frac{1}{2}) = \frac{MB}{2}

...(i)
The torque on the needle is

τ = M \times B

In magnitude,

τ = MB sin θ = MB sin 6 0^{\circ} = \frac{3}{2} MB

...(ii)
Dividing (ii) by (i), we get

\frac{τ}{W} = 3

τ = 3 W = 3 \times 3 J = 3 J

Q. A magnetic needle suspended parallel to a magnetic field requires $3 J$ of work to turn it through $6 0^{\circ} .$ The torque needed to maintain the needle in this position will be

$23 J$

$3 J$

$3 J$

$\frac{3}{2} J$

Q. A magnetic needle suspended parallel to a magnetic field requires 3​J of work to turn it through 60∘. The torque needed to maintain the needle in this position will be

23​J

3​J

3J

23​J

Q. A magnetic needle suspended parallel to a magnetic field requires $3 J$ of work to turn it through $6 0^{\circ} .$ The torque needed to maintain the needle in this position will be

$23 J$

$3 J$

$3 J$

$\frac{3}{2} J$