| (I) \[{{\vec{B}}_{1}}=3\hat{i}+7\hat{j}-5t\hat{k}\] |
| (II) \[{{\vec{B}}_{2}}=5t\hat{i}+4\hat{j}-15\hat{k}\] |
| (III) \[{{\vec{B}}_{3}}=2\hat{i}+5t\hat{j}-12\hat{k}\] |
| Where B is in millitesia and t is in second if the induced current in the loop due to \[{{\vec{B}}_{1}},\,{{\vec{B}}_{2}}\] and \[{{\vec{B}}_{3}}\] are \[{{i}_{1}},{{i}_{2}}\] and \[{{i}_{3}}\] respectively, then |
A) \[{{i}_{1}}>{{i}_{2}}>{{i}_{3}}\]
B) \[{{i}_{2}}>{{i}_{1}}>{{i}_{3}}\]
C) \[{{i}_{3}}>{{i}_{2}}>{{i}_{1}}\]
D) \[{{i}_{1}}={{i}_{2}}={{i}_{3}}\]
Correct Answer: B
Solution :
[b] \[{{i}_{1}}=\frac{d\phi }{dt}\frac{1}{R}=\frac{1}{R}\times \frac{d}{dt}\,\left( {{B}_{1}}\times Area \right)\] \[=\frac{d}{dt}\left[ 5t\times \frac{\pi {{a}^{2}}}{4} \right]\times \frac{1}{R}=\frac{5\pi {{a}^{2}}}{4R}\] \[{{i}_{2}}=\frac{d\phi }{dt}\frac{1}{R}=\frac{d\left[ 5{{a}^{2}}t \right]}{dt}\times \frac{1}{R}=\frac{5{{a}^{2}}}{R}\] \[{{i}_{3}}=\frac{d\phi }{dt}\frac{1}{R}=\frac{d\left[ 0.5{{a}^{2}} \right]}{dt}\times \frac{1}{R}=\frac{{{a}^{2}}}{R}\]
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