A bar magnet is held perpendicular to a uniform field. If the couple acting on the magnet is to be halved by rotating it, the angle by which it is to be rotated is

Given that θ1=90∘∴Initial torque τ1=τFinal torque τ2=τ/2 Now τ=τ1=MBsin⁡θ1=MBsin⁡90∘=MB  and τ2=MBsin⁡θ2 or τ/2=MBsin⁡θ2 ∴MB2=MBsin⁡θ2 or sin⁡θ2=12 ∴θ2=30∘So, the angle by which the bar magnet is to be rotated θ2=30o. dθ=θ1-θ2=900-300=600