A) \[48\text{ }V\]
B) \[4.8\text{ }V\]
C) \[4.8\times {{10}^{-1}}V\]
D) \[4.8\times {{10}^{-2}}V\]
Correct Answer: D
Solution :
From Faradays law of electromagnetic induction, the emf induced is given by \[e=-M\frac{di}{dt}\] ...(i) where M is coefficient of mutual inductance, \[\frac{di}{dt}\]is rate of change of current. Also, mutual inductance of two coaxial solenoids is given by \[M=\frac{{{\mu }_{0}}{{N}_{1}}{{N}_{2}}A}{l}\] ...(ii) From Eqs. (i) and (ii), we get \[e=\frac{{{\mu }_{0}}{{N}_{1}}{{N}_{2}}A}{l}\times \frac{di}{dt}\] Given, \[{{N}_{1}}=2000,{{N}_{2}}=300,\] \[A=1.2\times {{10}^{-3}}{{m}^{2}}.\] \[\frac{di}{dt}=\frac{2-(-2)}{0.25}=\frac{4}{0.25}\] \[\therefore \]\[e=\frac{4\pi \times {{10}^{-7}}\times 2000\times 300\times 1.2\times {{10}^{-3}}\times 4}{0.3\times 0.25}\] \[\Rightarrow \]\[e=\frac{4\times 3.14\times 2\times 3\times 1.2\times 4\times {{10}^{-5}}}{0.3\times 0.25}\] \[\Rightarrow \] \[|e|=4.8\times {{10}^{-2}}V\]You need to login to perform this action.
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