A circular coil having 200 turns, 2.5 x 10^ {-4} m^2 area and carrying 100 μA current

Question

A circular coil having 200 turns, \(2.5 \times 10^{-4} \mathrm{~m}^{2}\) area and carrying \(100 \ \mu \mathrm{A}\) current is placed in a uniform magnetic field of \(1 \mathrm{~T}\). Initially the magnetic dipole moment \((\overrightarrow{\mathrm{M}})\) was directed along \(\overrightarrow{\mathrm{B}}\). Amount of work, required to rotate the coil through \(90^{\circ}\) from its initial orientation such that \(\overrightarrow{\mathrm{M}}\) becomes perpendicular to \(\vec{B}\), is ________ \(\mu \mathrm{J}\).

Answer 1

Correct answer is : 5

\( \text { work done by external }=\Delta \mathrm{U}=\mathrm{U}_{\mathrm{f}}-\mathrm{U}_{\mathrm{i}}=\mathrm{M} . \mathrm{B} \)

\(\text { initially, } \mathrm{U}_{\mathrm{i}}=-\mathrm{MB} \cos \theta \)

\(=-\mathrm{MB}\)

\(\text { finally, } \mathrm{U}_{\mathrm{f}}=-\mathrm{Mb} \cos 90=0\)  

\(\mathrm{M}(\text { magnetic moment of coil })=\mathrm{i} . \mathrm{N} . \mathrm{A} \)  

\(\left(100 \times 10^{-6}\right) \cdot(200) \cdot\left(2.5 \times 10^{-4}\right)\)  

\(=5 \times 10^{-6}\)

\(\text { work }=\Delta \mathrm{U}=\mathrm{MB}=\left(5 \times 10^{-6}\right)(1) \)  

\(=5 \mu \mathrm{J}\)