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JEE Main & Advanced Physics Electro Magnetic Induction Question Bank Static EMI

  • question_answer
    A solenoid has 2000 turns wound over a length of 0.30 metre. The area of its cross-section is \[1.2\times {{10}^{-3}}{{m}^{2}}\]. 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 sec, then the e.m.f. induced in the coil is [NCERT 1982; MP PMT 2003]

    A)            \[6\times {{10}^{-4}}V\]  

    B)            \[4.8\times {{10}^{-3}}V\]

    C)            \[6\times {{10}^{-2}}V\]  

    D)            48 mV

    Correct Answer: D

    Solution :

                       Induced emf \[e=M\frac{di}{dt}=\frac{{{\mu }_{0}}{{N}_{1}}{{N}_{2}}A}{l}.\frac{di}{dt}\]                    \[=\frac{4\pi \times {{10}^{-7}}\times 2000\times 300\times 1.2\times {{10}^{-3}}}{0.30}\times \frac{|2-(-2)|}{0.25}\]            \[=48.2\times {{10}^{-3}}V=48\ mV\]


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