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A flexible circular loop `20cm` in diameter lies in a magneic field with magnitude `1.0 T`, direction lies into the plane of the page as shown in Fig. 3.187. The loop is pulled at the points indicated by the arrws, forming a loop of zero area in `0.314s`. if `R = 0,01 Omega`, the magnitude and direction of current flowing in the loop areA. (a) `1A`, clockwiseB. (b) `1A`, anticlockwiseC. ( c) `10 A`, clockwiseD. (d) `10 A`, anticlockwise |
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Answer» Correct Answer - C (c ) `epsilon_(av) = -(DeltaPhi_(B))/(Deltat) = -B(DeltaA)/(Deltat) = -B((-pir)^(2))/(Deltat)` Since the flux through the loop is decreasing, the induced current must produce a field that goes into the page. Therefore, the current flows in clockwise direction `I_(av) = (epsilon_(av) )/(R ) = (0.1)/(0.01) = 10 A` |
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