A) \[{{10}^{-4}}A\](Clockwise), \[2\times {{10}^{-4}}\,A\](Clockwise)
B) \[{{10}^{-4}}A\](Anticlockwise),\[2\times {{10}^{-4}}\,A\](Clockwise)
C) \[2\times {{10}^{-4}}\,A\](clockwise),\[{{10}^{-4}}\,A\](anticlockwise)
D) \[2\times {{10}^{-4}}\,A\](anticlockwise),\[{{10}^{-4}}A\](anticlockwise)
Correct Answer: A
Solution :
[a] Current in the inner oil \[i=\frac{\varepsilon }{R}=\frac{{{A}_{1}}}{{{R}_{1}}}\frac{dB}{dt}\] Length of the inner coil \[=2\pi a\] So its resistance \[{{R}_{1}}=50\times {{10}^{-3}}\times 2\pi (l)\] \[\therefore {{i}_{l}}=\frac{\pi {{a}^{2}}}{50\times {{10}^{-3}}\times 2\pi (\alpha )}\times 0.1\times {{10}^{-3}}={{10}^{-4}}\]According to lens?s law, direction of \[{{i}_{l}}\]is clockwise. Induced current in outer coil \[{{i}_{2}}=\frac{{{\varepsilon }_{2}}}{{{R}_{2}}}=\frac{{{A}_{2}}}{{{R}_{2}}}\frac{dB}{dt}\] \[\Rightarrow {{i}_{2}}=\frac{\pi {{b}^{2}}}{50\times {{10}^{-3}}\times (2\pi b)}\times 0.1\times {{10}^{-3}}=2\times {{10}^{-4}}(CW)\]
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