Q.A long solenoid $S$ has $n$ turns per metre, with diameter $a$. At the centre of this coil, we place a smaller coil of $N$ turns and diameter $b$ (where $b < a$). If the current in the solenoid increases linearly with time, what is the induced emf appearing in the smaller coil? Plot a graph showing the nature of variation in emf, if the current varies as a function of $mt^2 + C$.

Concept: Mutual Inductance — the emf induced in the smaller coil is due to the changing current (and hence changing flux) in the long solenoid.

Setup: the solenoid's field is $B = \mu_0 n i$, uniform over the smaller coil's cross-section (diameter $b < a$), so the flux linked is $\Phi = N\mu_0 n i \cdot \frac{\pi b^2}{4} = Mi$, giving mutual inductance $M = \dfrac{\mu_0 \pi N n b^2}{4}$. The induced emf is $\mathcal{E} = -M\dfrac{di}{dt}$.

Case 1 — current increases linearly ($i = kt$): $\dfrac{di}{dt} = k$ is constant, so the induced emf is constant — $\mathcal{E} = Mk$, independent of time. …

Why Mutual Inductance is the Right Lens

When current changes in the solenoid, the magnetic field inside it changes. That changing field passes through the smaller coil placed at the centre, inducing an emf in it. This is a textbook case of mutual inductance — the two coils are magnetically coupled, and the emf in the secondary (small coil) depends only on how fast the current in the primary (solenoid) changes.

The beauty here is that the solenoid's field is uniform inside (for an ideal long solenoid), so the flux through the small coil is simply $B$ times its area. No messy integration over position.

Step-by-Step Solution

1. Magnetic field of the solenoid

For an ideal long solenoid with $n$ turns per metre carrying current $i$, the magnetic field inside is uniform and axial:

This field is constant over the cross-section of the solenoid. Since the small coil has diameter $b < a$, it lies entirely within this uniform field.

2. Flux through the small coil

The small coil has $N$ turns, each of area $\pi (b/2)^2 = \frac{\pi b^2}{4}$. The flux through one turn is $B \times \text{area}$, and through all $N$ turns:

So:

The quantity in front of $i$ is the mutual inductance $M$:

3. Induced emf from Faraday's law

The induced emf in the small coil is:

We only care about magnitude (direction is given by Lenz's law, but the problem asks for the emf value, so we take magnitude unless sign is requested).

4. Case 1: Current increases linearly with time

If $i = kt$ (where $k$ is a constant), then $\frac{di}{dt} = k$. So:

This is constant — independent of time.

5. Case 2: Current varies as $i = m t^2 + C$

Here $m$ and $C$ are constants. Differentiate:

So the induced emf becomes:

That is: …

Method: Mutual Inductance from Field Linkage, Then Differentiating an Arbitrary Current Waveform

Use this two-part technique whenever a question gives a primary coil/solenoid and asks for the emf induced in a nearby secondary coil, especially when the primary's current is described by different functions of time.

Steps

Step 1: Find the primary's field at the secondary's location

For a long solenoid, the field inside is uniform: $B=\mu_0 n i$. Check whether the secondary coil lies entirely within this uniform region (its diameter smaller than the solenoid's) — if so, no spatial integration is needed.

Step 2: Find the flux linked by the secondary

Step 3: Extract the mutual inductance M

Write $\Phi=Mi$ and read off $M$ as the coefficient multiplying $i$ — this is purely a geometric quantity (turns, areas, separation), fixed once and for all regardless of how $i$ varies in time.

Step 4: Apply the general emf formula

This single formula, once $M$ is known, applies to any current waveform $i(t)$ — you never need to redo Steps 1–3 for a different $i(t)$.

Step 5: Differentiate the given $i(t)$ …