A coil having n turns and resistance R Omega | vedclass

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(B) Given: Number of turns in the coil $= n$. Resistance of the coil $= R$. Resistance of the galvanometer $= 4R$. Total resistance of the circuit $R_

Vedclass · Accepted Answer

$-\frac{n(\phi_2-\phi_1)}{5 Rt}$

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(B) Given: Number of turns in the coil $= n$. Resistance of the coil $= R$. Resistance of the galvanometer $= 4R$. Total resistance of the circuit $R_{total} = R + 4R = 5R$.According to Faraday's Law of electromagnetic induction,the induced electromotive force (emf) is given by $e = -n \frac{\Delta \phi}{\Delta t}$.Here,the change in flux is $\Delta \phi = \phi_2 - \phi_1$ and the time taken is $\Delta t = t$.Therefore,$e = -n \frac{(\phi_2 - \phi_1)}{t}$.The induced current $i$ is given by $i = \frac{e}{R_{total}}$.Substituting the values,we get $i = \frac{-n(\phi_2 - \phi_1)}{5Rt}$.

$A$ coil having $n$ turns and resistance $R \ \Omega$ is connected with a galvanometer of resistance $4 R \ \Omega$. This combination is moved in time $t$ seconds from a magnetic flux $\phi_1$ Weber to $\phi_2$ Weber. The induced current in the circuit is

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