Question 123: Magnetism and Matter

Question

A magnetic needle lying parallel to a magnetic field requires $$W$$ units of work to turn it through $${60^ \circ }.$$ The torque needed to maintain the needle in this position will be :

Accepted Answer

$$W = MB\left( {\cos {	heta _1} - \cos {	heta _2}} ight)$$
$$ = MB\left( {\cos {	heta ^ \circ } - \cos {{60}^ \circ }} ight)$$
$$ = MB\left( {1 - {1 \over 2}} ight) = {{MB} \over 2}$$
$$	herefore$$ $$	au  = MB\,\sin 	heta  = MB\,\sin \,{60^ \circ }$$
$$ = \sqrt 3 {{MB} \over 2} = \sqrt 3 W$$

Answer

$$\sqrt 3 \,W$$

Answer

$$W$$

Answer

$${{\sqrt 3 } \over 2}W$$

Answer

$$2W$$

Magnetism and Matter

A magnetic needle lying parallel to a magnetic field requires $W units of work to turn it through {60^ \circ }.$ The torque needed to maintain the needle in this position will be :

Choose the correct answer: