JEE Main 2019 — Magnetic Properties of Matter Question with Solution

As we know that, work done in rotating a magnetic dipole in magnetic field is,

$W=MB(\cos \left({\theta }_1\right)-\cos \left({\theta }_2\right)) ...\left(1\right)$

Now in the above given question we have, Magnetic Moment $\left(M\right)={10}^{-2} \hat{\mathrm{i}} \mathrm{A} {\mathrm{m}}^2$,Magnetic Field$=B\hat{\mathrm{i}}\cos \left(\omega t\right)$ with $B=1 \mathrm{T}$ , $\omega =0.125 \mathrm{rad} {\mathrm{s}}^{-1}$ and $t=1 \mathrm{s}$

Let initial angle $\left({\theta }_1\right)=0^{\circ }$ so on reversing the direction of magnetic moment final angle $\left({\theta }_2\right)={180}^{\circ }$

Now, substituting all the values in equation $\left(1\right)$ we get,

$\Rightarrow W={10}^{-2}\times \left(1\right)\cos \left(0.125\times 1\right)\left(\cos \left(0^{\circ }\right)-\cos \left({180}^{\circ }\right)\right)$

$\Rightarrow W={10}^{-2}\cos \left(0.125\right)\left(1-\left(-1\right)\right) (\because \cos \left(0^{\circ }\right)=1 \text{and} \cos \left({180}^{\circ }\right)=-1)$

$\Rightarrow W=2\times {10}^{-2}\times 0.99$

$\Rightarrow W\approx 0.02 \mathrm{J}$

Therefore, the work done in reversing the direction of the magnetic moment is $0.02 \mathrm{J}$.