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A long solenoid with 10 turns per cm has a small loop of area \mathrm{3 \; cm^{2}} placed inside, normal to the axis of the solenoid. If the current carried by the solenoid changes steadily from 2 A to 4 A in 0.2 s, what is the induced voltage in the loop, while the current is changing?

Option: 1

\mathrm{4.2 \times 10^{-8} \mathrm{~V}}


Option: 2

\mathrm{2.8 \times 10^{-8} \mathrm{~V}}


Option: 3

\mathrm{7.3 \times 10^{-6} \mathrm{~V}}


Option: 4

\mathrm{3.8 \times 10^{-6} \mathrm{~V}}


Answers (1)

best_answer

\text { Here, } \frac{\mathrm{N}}{l}=10 \text { turns per } \mathrm{cm}=1000 \text { turns per } \mathrm{m}

\begin{aligned} & \mathrm{A}=3 \mathrm{~cm}^2-3 \times 10^{-4} \mathrm{~m}^2 \\ & \text { or } \quad \frac{\mathrm{dI}}{\mathrm{dt}}=\frac{4-2}{0.2}=10 \mathrm{As}^{-1} \end{aligned}

\text { Also } \varepsilon=\frac{\mathrm{d} \phi}{\mathrm{dt}}=\frac{\mathrm{d}}{\mathrm{dt}}(\mathrm{BA})=\mathrm{A} \frac{\mathrm{d}}{\mathrm{dt}}\left(\mu_0 \frac{\mathrm{N}}{l} \mathrm{I}\right)=\mathrm{A} \mu_0\left(\frac{\mathrm{N}}{l}\right) \frac{\mathrm{dI}}{\mathrm{dt}}

=-3 \times 10^{-4} \times 4 \pi \times 10^{-7} \times 1000 \times 10-3.8 \times 10^{-6} \mathrm{~V}

Posted by

Ajit Kumar Dubey

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