Question

A coil having n turns \(\&\) resistance \(R \Omega\) is connected with a galvanometer of resistance \(4 \mathrm{R} \Omega\). This combination is moved from a magnetic field \(\mathrm{W}_{1} \mathrm{~Wb}\) to $\mathrm{W}_{2} \mathrm{~Wb}\( in \)\mathrm{t}$ second. The induced current in the circuit is.... (a) $-\left[\left\\{\mathrm{W}_{2}-\mathrm{W}_{1}\right\\} /\\{5 \mathrm{Rnt}\\}\right]$ (b) $-\mathrm{n}\left[\left\\{\mathrm{W}_{2}-\mathrm{W}_{1}\right\\} /\\{5 \mathrm{Rt}\\}\right]$ (c) $-\left[\left\\{\mathrm{W}_{2}-\mathrm{W}_{1}\right\\} /\\{\mathrm{Rnt}\\}\right]$ (d) $-\mathrm{n}\left[\left\\{\mathrm{W}_{2}-\mathrm{W}_{1}\right\\} /\\{\mathrm{Rt}\\}\right]$

Short answer

The short answer is: \(I = -n\left[\frac{W_2-W_1}{Rt}\right]\).

Step-by-step solution

Calculate the change in magnetic flux

First, find the change in magnetic flux. The magnetic flux can be calculated as the product of the number of turns \(n\) and the change in magnetic field \(\Delta W = W_2 - W_1\): \[\Delta \Phi = n\Delta W\]

Apply Faraday's law of electromagnetic induction

According to Faraday's law of electromagnetic induction, the induced electromotive force (EMF) in the circuit is equal to the negative rate of change of magnetic flux: \[EMF = -\frac{d\Phi}{dt}\] Since the change in magnetic flux happens in \(t\) seconds, we replace \(d\Phi\) with \(\Delta \Phi\) and \(dt\) with \(t\): \[EMF = -\frac{\Delta \Phi}{t}\]

Substitute the values from Step 1 into the equation from Step 2

Substitute the values from Step 1 into the equation from Step 2: \[EMF = -\frac{n\Delta W}{t}\]

Apply Ohm's law to find the induced current

According to Ohm's law, the induced current in the circuit is equal to the induced EMF divided by the total resistance in the circuit: \[I = \frac{EMF}{R_t}\] The total resistance in the circuit is the sum of the resistance of the coil and the galvanometer: \[R_t = R + 4R = 5R\]

Substitute the values from Steps 3 and 4 to find the induced current

Substitute the values from Steps 3 and 4 into the equation for the induced current: \[I = \frac{-n\Delta W}{5Rt}\] We can rewrite \(\Delta W\) as \(W_2 - W_1\): \[I = -\frac{n(W_2-W_1)}{5Rt}\] This expression matches option (d) among the given choices. Thus, the induced current in the circuit is: \[I = -n\left[\frac{W_2-W_1}{Rt}\right]\]