A coil having n turns and resistance R ohm is connected with a galvanometer of resistance 4R ohm. Th

Q: Why is the total resistance $5R$ and not just $R$?

The coil itself has an internal resistance of $R$, and it is connected in series with a galvanometer that has a resistance of $4R$. In a series circuit, resistances add up, so the total equivalent resistance is $R + 4R = 5R$.

Q: The question says 'magnetic field $W_1$ weber', isn't weber the unit of flux?

You are absolutely right! The phrasing in the question is a bit loose. However, the unit 'weber' definitively tells us that $W_1$ and $W_2$ are values of magnetic flux ($\Phi$), not the magnetic field ($B$).

Q: Why is there a negative sign in the final answer?

The negative sign comes directly from Faraday's Law ($e = -n \frac{d\Phi}{dt}$), which incorporates Lenz's Law. It signifies that the induced current will flow in a direction that opposes the change in magnetic flux.

Q: Do we need to multiply by $n$ for the total flux?

Yes. The values $W_1$ and $W_2$ represent the flux linked with a single turn. For a coil with $n$ turns, the total effective flux linkage is $n\Phi$, which is why Faraday's formula includes the factor $n$.

Q: What if the galvanometer was connected in parallel?

If it were in parallel, the equivalent resistance would be calculated using the parallel formula $(R \cdot 4R)/(R + 4R) = 0.8R$. However, the problem describes a single continuous loop, which implies a series connection.

Question

A coil having $n$ turns and resistance $R\ \Omega$ is connected with a galvanometer of resistance $4R\ \Omega$. This combination is moved for time $t$ seconds from a magnetic field $W_1$ weber to $W_2$ weber. The induced current in the circuit is

Options:

(A) $-\frac{W_2 - W_1}{5Rnt}$

(B) $-\frac{n(W_2 - W_1)}{5Rt}$

(C) $-\frac{(W_2 - W_1)}{Rnt}$

(D) $-\frac{n(W_2 - W_1)}{Rt}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are in a lab, looking at a simple yet fascinating electrical circuit. We have a coil, which is essentially a tightly wound wire with $n$ turns. This coil isn't perfect; it has its own internal resistance of $R\ \Omega$.

Connected in series with this coil is a galvanometer. A galvanometer is a sensitive device used to detect small currents, and in our setup, it has a resistance of $4R\ \Omega$.

Because these two components are connected end-to-end in a single loop, they are in series. **The total equivalent resistance** of our circuit is simply the sum of their individual resistances. So, $R_{eq} = R + 4R = 5R$. This $5R$ is the total opposition any induced current will face.

### Decoding the Magnetic Flux

Now, the problem throws a slight curveball. It says the "magnetic field" changes from $W_1$ weber to $W_2$ weber.

Here is where you need to be sharp! The unit given is **weber**, which is the standard unit for **magnetic flux ($\Phi$)**, not magnetic field intensity ($B$). This tells us that $W_1$ and $W_2$ represent the initial and final magnetic flux linked with each turn of the coil.

The change in this magnetic flux over a time interval of $t$ seconds is what drives the entire phenomenon. Mathematically, the change in flux is $d\Phi = W_2 - W_1$.

### The Master Equation: Faraday's Law

Whenever magnetic flux through a coil changes, nature reacts by inducing an electromotive force (EMF). This is beautifully captured by **Faraday's Law of Electromagnetic Induction**.

For a coil with $n$ turns, the induced EMF is given by $e = -n \frac{d\Phi}{dt}$. The negative sign is crucial—it represents Lenz's Law, indicating that the induced EMF will always oppose the change in flux that created it.

Substituting our specific values into Faraday's equation, we get the raw induced EMF: $e = -n \frac{W_2 - W_1}{t}$.

### Final Calculation: Finding the Current

We have the EMF pushing the electrons, and we have the total resistance trying to slow them down. To find the actual induced current $I$, we bring in our trusty old friend, **Ohm's Law**.

Ohm's Law states that $I = \frac{e}{R_{eq}}$.

Let's substitute the expressions we've derived. We plug in our EMF and our total resistance of $5R$. This gives us $I = \frac{-n \frac{W_2 - W_1}{t}}{5R}$.

By neatly rearranging the terms, we arrive at our final, elegant expression for the induced current: **$I = -\frac{n(W_2 - W_1)}{5Rt}$**. This perfectly matches option (b), completing our journey through the problem!

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Visualized Solution

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The Sigma Insight: Faraday's Laws of Electromagnetic Induction

Analyzing the Setup

Decoding the Magnetic Flux

The Master Equation

Final Calculation

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Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Circuit Setup

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Equivalent Resistance

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Change in Magnetic Flux

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Faraday′s Law

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Induced EMF

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Ohm′s Law

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Final Calculation