A long solenoid having 100 turns per $\mathrm{cm}$ carries a current of $\frac{4}{\pi}$ A. At the centre of it is placed a coil of 200 turns of cross-sectional area $25 \mathrm{~cm}^2$ having its axis parallel to the field produced by the solenoid. When the direction of the current in the solenoid…

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A long solenoid having 100 turns per $\mathrm{cm}$ carries a current of $\frac{4}{\pi}$ A. At the centre of it is placed a coil of 200 turns of cross-sectional area $25 \mathrm{~cm}^2$ having its axis parallel to the field produced by the solenoid. When the direction of the current in the solenoid is reversed with in $0.04 \mathrm{~s}$, the induced emf in the coil is

Aicharya · Accepted Answer

Due to solenoid a magnetic field exists around coil placed at centre

Magnetic field intensity produced by solenoid, $B=\mu_0 n I$ Flux linked with coil is $\phi_B=N . B \cdot A$ Where, $N=$ Number of turns in coil $A=$ Area of coil As current in solenoid is reversed, change in flux is $\Delta \phi_B=2 N \cdot B \cdot A$ This change in flux produces an emf in coil given by Induced emf $e=$ Rate of change of flux $\Rightarrow \quad e=\frac{2 N B A}{\Delta t}=\frac{2 N\left(\mu_0 n I\right) A}{\Delta t}$ Here $N=200, A=25 \mathrm{~cm}^2=25 \times 10^{-4} \mathrm{~m}^2$ $I=\frac{4}{\pi} \mathrm{A}, n=100$ tuns $/ \mathrm{cm}=100 \times 100$ turns $/ \mathrm{m}$ $\mu_0=4 \pi \times 10^{-7} \mathrm{~T} \cdot \mathrm{mA}^{-1}$ and $\Delta t=0.04 \mathrm{~s}$ So induced emf, $e=\frac{2 \times 200 \times 4 \pi \times 10^{-7} \times 10^4 \times \frac{4}{\pi} \times 25 \times 10^{-4}}{0.04}=0.4 \mathrm{~V}$ Induced emf in coil is $0.4 \mathrm{~V}$

A long solenoid having 100 turns per $\mathrm{cm}$ carries a current of $\frac{4}{\pi}$ A. At the centre of…

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