A) 0.3 V
B) 9.6 V
C) 30.0 V
D) 3.0 V
Correct Answer: A
Solution :
[a] \[{{e}_{back}}=L\frac{di}{dt}\] where \[L=\frac{{{\mu }_{0}}{{N}^{2}}A}{l}\] \[\therefore \,\,\,\,\,{{e}_{back}}=\frac{{{\mu }_{0}}{{N}^{2}}A}{l}\left[ \frac{1.5-0}{1\times {{10}^{3}}} \right]\] \[=\frac{4\pi \times {{10}^{-7}}{{(400)}^{2}}\times 20\times {{10}^{-4}}}{20\times {{10}^{-2}}}\times (1.5\times {{10}^{3}})\] = 0.3 V
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