Get Answers to all your Questions

header-bg qa

A bar magnet having a magnetic movement of 2×10-4 JT-1 is free to rotate in a horizontal plane. A horizontal magnetic field B=6×10-4 T exists in the space. The work done in taking the magnet slowly from a direction parallel to the field to a direction 60o from the field is :

Answers (1)

\\\text{The initial direction of magnet = parallel to the field }=0^{\circ}\\ \text{The final direction of the magnet }=60^{\circ}\\ \text{Magnetic field, } \mathrm{B}=6 \times 10^{-4} \mathrm{T}\\ \text{Magnetic Moment, }\mathrm{M}=2 \times 10^{4} \mathrm{JT}^{-1}

\\\text{Work done in rotating a magnet in the magnetic field is given by:} \\ \mathrm{W}=\mathrm{B} \cdot \mathrm{T}\left(\cos \theta_{1}-\cos \theta_{2}\right)\\ \mathrm{W}=6 \times 10^{-4} \times 2 \times 10^{4}(1-0.5)\\ \mathrm{W}=6 \mathrm{J} \text{Thus, It requires a } 6 \mathrm{J}\\ \text{of work to rotate the magnet from } 0^{\circ} \ to \ 60^{\circ} \text{in the magnetic field.}

Posted by

lovekush

View full answer