question_answer

A solenoid has 2000 turns wound over a length of 0.30 m. The area of its cross-section is \[1.2\times {{10}^{-3}}\text{ }{{m}^{2}}.\] Around its central section a coil of 300 turns is wound. If an initial current of 2 A in the solenoid is reversed in 0.25 s, the emf induced in the coil is:

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, \[{{N}_{1}}=2000,{{N}_{2}}=300,\] \[A=1.2\times {{10}^{-3}}{{m}^{2}}.\] \[\frac{di}{dt}=\frac{2-(-2)}{0.25}=\frac{4}{0.25}\] \[\therefore \]\[e=\frac{4\pi \times {{10}^{-7}}\times 2000\times 300\times 1.2\times {{10}^{-3}}\times 4}{0.3\times 0.25}\] \[\Rightarrow \]\[e=\frac{4\times 3.14\times 2\times 3\times 1.2\times 4\times {{10}^{-5}}}{0.3\times 0.25}\] \[\Rightarrow \]               \[|e|=4.8\times {{10}^{-2}}V\]