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.

Answer:

(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 \[\theta =90{}^\circ \] \[\therefore \]      \[\phi =BA\cos 90{}^\circ =0\]