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A long solenoid S has n turns per meter, with a radius of a. At the center of this coil, we place a smaller coil of N turns and radius b (where b<a). If the current in the solenoid increases linearly with time, what is the induced emf appearing in the smaller coil? Plot a graph showing the nature of variation in emf, if current varies as a function of mt^{2}+C.

Answers (1)

Explanation:-

The magnetic field caused by a solenoid is given by

B=\mu _{0}ni.

Magnetic flux in the smaller coil is

\phi _{m}=NBA

where, A=\pi b^{2}

Applying Faraday's law of EMI, we have

So, e=\frac{-d\phi }{dt}=\frac{-d}{dt}(NBA)

           =-N\pi b^{2}\frac{d(B)}{dt}

Where, B=\mu _{0}Ni

\Rightarrow e=-N\pi b^{2}\; \mu _{0}n\frac{di}{dt}

Since, current varies as a function of time, so

i(t)=mt^{2}+C

\Rightarrow \; e=-Nn\pi \mu _{0}b^{2}\frac{d}{dt}(mt^{2}+C) 

    

By solvinge=-\mu _{0}Nn\pi b^{2}2mt

The negative sign signifies the opposite nature of induced emf.

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infoexpert23

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