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QJEEM24Ph25a7: JEE Main 2024 Physics - Class 12 JEEM24Ph25a7 Solution: A circular coil having 200 turns, area and carrying current is…

A circular coil having 200 turns, 2.5×10−4 m22.5 \times 10^{-4} \mathrm{~m}^2 area and carrying 100μA100 \mu \mathrm{A} current is placed in a uniform magnetic field of 1 T1 \mathrm{~T} . Initially the magnetic dipole moment (M⃗)(\vec{M}) was directed along B⃗\vec{B} . Amount of work, required to rotate the coil through 90∘90^{\circ} from its initial orientation such that M⃗\vec{M} becomes perpendicular to B⃗\vec{B} , is ________ μ\mu J.

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amp;W=Uf−Ui=(−MBcos⁡90)−(−MBcos⁡0)amp;⇒W=MB=NiAB=5μJ\begin{aligned}& W=U_f-U_i=(-M B \cos 90)-(-M B \cos 0) \\& \Rightarrow W=M B=N i A B=5 \mu \mathrm{J}\end{aligned}

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