A) 10? 4 A (Clockwise), 2 ´ 10? 4 A (Clockwise)
B) 10? 4 A (Anticlockwise), 2 ´ 10? 4 A (Clockwise)
C) 2 ´ 10? 4 A (clockwise), 10? 4 A (Anticlockwise)
D) 2 ´ 10? 4 A (Anticlockwise), 10? 4 A (Anticlockwise)
Correct Answer: A
Solution :
Current in the inner coil \[i=\frac{e}{R}=\frac{{{A}_{1}}}{{{R}_{1}}}\frac{dB}{dt}\] length of the inner coil \[=2\pi a\] so it?s resistance \[{{R}_{1}}=50\times {{10}^{-3}}\times 2\pi \ (a)\] \[\therefore {{i}_{1}}=\frac{\pi {{a}^{2}}}{50\times {{10}^{-3}}\times 2\pi \ (a)}\times 0.1\times {{10}^{-3}}={{10}^{-4}}A\] According to lenz?s law direction of i1 is clockwise. Induced current in outer coil \[{{i}_{2}}=\frac{{{e}_{2}}}{{{R}_{2}}}=\frac{{{A}_{2}}}{{{R}_{2}}}\frac{dB}{dt}\] \[\Rightarrow {{i}_{2}}=\frac{\pi {{b}^{2}}}{50\times {{10}^{-3}}\times (2\pi b)}\times 0.1\times {{10}^{-3}}=2\times {{10}^{-4}}A\ (CW)\]
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