Magnetic Properties of Matter

JEE Main level, single correct, medium

21 of 57

Question 21 of 57

A magnet of total magnetic moment 10−2i^Am2{10}^{-2}\hat{i}A{m}^{2} is placed in a time varying magnetic field, Bi^(cos⁡ωt)B\hat{i}\left(\cos \omega t\right) where B=1B=1 Tesla and ω=0.125rads−1\omega =0.125rad{s}^{-1}. The work done for reversing the direction of the magnetic moment at t=1t=1 second, is:

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Answer: (B) 0.02J0.02J

As we know that, work done in rotating a magnetic dipole in magnetic field is,
W=MB(cos⁡(θ1)−cos⁡(θ2))...(1)W=MB(\cos \left({\theta }_{1}\right)-\cos \left({\theta }_{2}\right))...(1)
Now in the above given question we have, Magnetic Moment (M)=10−2i^Am2(M)={10}^{-2}\hat{i}A{m}^{2},Magnetic Field=Bi^cos⁡(ωt)=B\hat{i}\cos \left(\omega t\right) with B=1TB=1T , ω=0.125rads−1\omega =0.125rad{s}^{-1} and t=1st=1s
Let initial angle (θ1)=0∘({\theta }_{1})={0}^{^{\circ}} so on reversing the direction of magnetic moment final angle (θ2)=180∘({\theta }_{2})={180}^{^{\circ}}
Now, substituting all the values in equation (1)(1) we get,
⇒W=10−2×(1)cos⁡(0.125×1)(cos⁡(0∘)−cos⁡(180∘))\Rightarrow W={10}^{-2}\times (1)\cos \left(0.125\times 1\right)\left(\cos \left({0}^{^{\circ}}\right)-\cos \left({180}^{^{\circ}}\right)\right)
⇒W=10−2cos⁡(0.125)(1−(−1))(∵cos⁡(0∘)=1andcos⁡(180∘)=−1)\Rightarrow W={10}^{-2}\cos \left(0.125\right)\left(1-\left(-1\right)\right)(∵\cos \left({0}^{^{\circ}}\right)=1\text{and}\cos \left({180}^{^{\circ}}\right)=-1)
⇒W=2×10−2×0.99\Rightarrow W=2\times {10}^{-2}\times 0.99
⇒W≈0.02J\Rightarrow W\approx 0.02J
Therefore, the work done in reversing the direction of the magnetic moment is 0.02J0.02J.

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