• # question_answer (i) A metallic piece becomes very hot when it is surrounded by a coil carrying high frequency alternating current. (ii) A rectangular loop of area $20\,cm\times 30\,cm$ is placed in magnetic field of 0.3T, with its plane (a) inclined $30{}^\circ$ to the field. (b) parallel to the field. Find out the magnetic flux linked with the loop.

(i) Due to flow of high frequency alternating current in the coil, the magnetic flux linked with the metallic piece changes by a large amount, so heavy eddy currents are induced in the metallic piece, causes heating. (ii) Given,$~A=20\times 30=600\,c{{m}^{2}}$ $=6\times {{10}^{-2}}{{m}^{2}},$ $B=0.3T$ Let $\theta$ be the angle between magnetic field and the area vector of the loop. (a) Here, we have             $\theta =90{}^\circ -30{}^\circ =60{}^\circ$ $\therefore$ Magnetic flux, $\phi =B.A=BA\,\,\cos \theta$             $=0.3\times 6\times {{10}^{-2}}\times \cos 60{}^\circ$ $=0.3\times 6\times {{10}^{-2}}\times \frac{1}{2}=0.9\times {{10}^{-2}}Wb$ (b) Here, we have $\theta =90{}^\circ$ $\therefore$      $\phi =BA\cos 90{}^\circ =0$