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 :

A thin rectangular magnet suspended freely has a period of oscillation equal to $$T.$$ Now it is broken into two equal halves (each having half of the original length) and one piece is made to oscillate freely in the same field. If its period of oscillation is $$T',$$ the ratio $${{T'} \over T}$$ is

The magnetic lines of force inside a bar magnet