Q.A long solenoid with $15$ turns per cm has a small loop of area $2.0\ \text{cm}^2$ placed inside the solenoid normal to its axis. If the current carried by the solenoid changes steadily from $2.0\ \text{A}$ to $4.0\ \text{A}$ in $0.1\ \text{s}$, what is the induced emf in the loop while the current is changing?

Concept: Mutual Inductance — a changing current in the solenoid produces a changing magnetic flux through the loop, inducing an emf.

Step 1: Magnetic field inside the solenoid

$B = \mu_0 n I$, where $n = 15$ turns/cm $= 1500$ turns/m.

Step 2: Flux through the loop

$\Phi = B A = \mu_0 n I A$, with $A = 2.0\ \text{cm}^2 = 2.0 \times 10^{-4}\ \text{m}^2$.

Step 3: Induced emf

$\mathcal{E} = -\frac{d\Phi}{dt} = -\mu_0 n A \frac{dI}{dt}$.

Here $\frac{dI}{dt} = \frac{4.0 - 2.0}{0.1} = 20\ \text{A/s}$.

Step 4: Substitute values

$\mu_0 = 4\pi \times 10^{-7}\ \text{T m/A}$, so

$\mathcal{E} = (4\pi \times 10^{-7})(1500)(2.0 \times 10^{-4})(20)$.

Compute:

$4\pi \times 10^{-7} \times 1500 = 6\pi \times 10^{-4}$

Multiply by $2.0 \times 10^{-4}$ gives $1.2\pi \times 10^{-7}$

Multiply by $20$ gives $2.4\pi \times 10^{-6}\ \text{V}$.

The key here is mutual inductance — the solenoid’s magnetic field links the small loop, and when the solenoid current changes, the flux through the loop changes, inducing an emf. You don’t need the mutual inductance coefficient explicitly; you can compute the flux directly because the field inside a long solenoid is uniform and given by $B = \mu_0 n I$, where $n$ is the number of turns per unit length.

Let’s work through it step by step.

1. Find the magnetic field inside the solenoid. For an ideal long solenoid, the field is uniform along the axis and given by

where $\mu_0 = 4\pi \times 10^{-7}\ \text{T m/A}$, $n$ is the number of turns per metre, and $I$ is the current.

Here, $n = 15\ \text{turns per cm} = 1500\ \text{turns per metre}$.

So at any instant, $B = (4\pi \times 10^{-7}) \times 1500 \times I = 6\pi \times 10^{-4} \times I$ tesla.

2. Compute the magnetic flux through the small loop. The loop is placed normal to the solenoid’s axis, so the field is perpendicular to its area. Flux is

where $A = 2.0\ \text{cm}^2 = 2.0 \times 10^{-4}\ \text{m}^2$.

Thus

in webers.

3. Find the rate of change of flux. The current changes steadily from $2.0\ \text{A}$ to $4.0\ \text{A}$ in $0.1\ \text{s}$, so

Since $\Phi$ is proportional to $I$,

4. Apply Faraday’s law. The induced emf in the loop is

The magnitude is

Method: Faraday's Law of Electromagnetic Induction (via Mutual Inductance)

We use the mutual inductance approach — the induced emf in the loop depends on the rate of change of current in the solenoid and the mutual inductance between them.

Steps

1. Find the number of turns per unit length of the solenoid

Given: $15$ turns per cm

Convert to SI units:

2. Magnetic field inside the solenoid

For an ideal long solenoid, the field inside is uniform and given by:

where $\mu_0 = 4\pi \times 10^{-7} \ \text{T m/A}$.

3. Magnetic flux through the small loop

The loop is placed normal to the axis, so the flux is:

Area $A = 2.0 \ \text{cm}^2 = 2.0 \times 10^{-4} \ \text{m}^2$

4. Induced emf from Faraday's Law

5. Calculate the rate of change of current

Current changes from $2.0 \ \text{A}$ to $4.0 \ \text{A}$ in $0.1 \ \text{s}$:

6. Substitute values

7. Simplify step-by-step

• $4\pi \times 10^{-7} \times 1500 = 6\pi \times 10^{-4}$
• $6\pi \times 10^{-4} \times 2.0 \times 10^{-4} = 12\pi \times 10^{-8}$
• $12\pi \times 10^{-8} \times 20 = 240\pi \times 10^{-8}$

8. Final result

(using $\pi \approx 3.14$)

The magnitude of the induced emf is $7.54 \ \mu\text{V}$.

Here are the common mistakes students make on this Mutual Inductance problem, and how to avoid each.

1. Forgetting to convert units correctly

The Mistake:

Using $15$ turns per cm directly as $n = 15$ in the formula $B = \mu_0 n I$, without converting to turns per metre.

Why it’s wrong:

The SI unit of $\mu_0$ is $\text{T m/A}$, so $n$ must be in turns per metre. Using turns per cm gives a result off by a factor of 100.

How to avoid:

Always write the conversion step explicitly:

$n = 15\ \text{turns/cm} = 15 \times 100 = 1500\ \text{turns/m}$.

2. Using the wrong formula for magnetic field inside a solenoid

The Mistake:

Using $B = \mu_0 n I$ for a finite solenoid or using $B = \mu_0 N I / L$ but confusing $N$ (total turns) with $n$ (turns per unit length).

Why it’s wrong:

For a long solenoid, the field is uniform and given by $B = \mu_0 n I$. If you use total turns $N$, you must also use the correct length $L$.

How to avoid:

• Identify that “long solenoid” means $B = \mu_0 n I$ is valid.
• If given turns per unit length, use $n$ directly.
• If given total turns $N$ and length $L$, use $n = N/L$.

3. Confusing area units

The Mistake:

Plugging $A = 2.0\ \text{cm}^2$ directly into the flux formula without converting to $\text{m}^2$.

Why it’s wrong:

$1\ \text{cm}^2 = 10^{-4}\ \text{m}^2$, so $2.0\ \text{cm}^2 = 2.0 \times 10^{-4}\ \text{m}^2$. Using cm² gives an emf that is 10,000 times too large.

How to avoid:

Convert all areas to $\text{m}^2$ before calculation:

$A = 2.0\ \text{cm}^2 = 2.0 \times 10^{-4}\ \text{m}^2$.

4. Misapplying Faraday’s law sign convention

The Mistake:

Writing $\mathcal{E} = -\frac{d\phi}{dt}$ and then reporting the emf as negative without stating the direction, or ignoring the sign entirely.

Why it’s wrong:

The question asks for “induced emf” — usually the magnitude is expected unless direction is specifically asked. A negative sign without explanation can lose marks.

How to avoid:

• If only magnitude is asked, give the absolute value: $|\mathcal{E}| = \left| -\frac{d\phi}{dt} \right| = \frac{d\phi}{dt}$.
• If direction is asked, use Lenz’s law separately.

5. Using the wrong time interval

The Mistake:

Using $\Delta t = 0.1\ \text{s}$ but taking the change in current as $4.0\ \text{A} - 2.0\ \text{A} = 2.0\ \text{A}$ correctly, but then dividing by the wrong time (e.g., using 0.1 s as the time for one turn).

Why it’s wrong:

The time interval is for the entire current change, not per turn.

How to avoid:

Write clearly:

$\frac{dI}{dt} = \frac{4.0 - 2.0}{0.1} = \frac{2.0}{0.1} = 20\ \text{A/s}$.

6. Forgetting that flux links the loop only once

The Mistake:

Multiplying the flux by the number of turns of the solenoid (1500) when calculating emf in the loop.

Why it’s wrong:

The small loop has only one turn. The solenoid’s turns create the field, but the induced emf is in the loop, not in the solenoid.

How to avoid:

• Flux through the loop: $\phi = B \cdot A$ (one turn).
• Induced emf: $\mathcal{E} = -\frac{d\phi}{dt}$ (no extra factor of $N$ for the loop).

7. Mixing up mutual inductance and self-inductance

The Mistake:

Using $M = \mu_0 n_1 n_2 A l$ or similar formula for mutual inductance, then calculating emf as $M \frac{dI}{dt}$, but getting the geometry wrong.

Why it’s wrong:

Here, the mutual inductance is simply $M = \mu_0 n A$ (for the loop inside the solenoid), but students often overcomplicate.

How to avoid:

• For a small loop inside a long solenoid: $M = \mu_0 n A$.
• Then $\mathcal{E} = M \frac{dI}{dt}$ directly.
• Or compute $B$, then $\phi$, then emf — both give the same answer.

Quick checklist to avoid all mistakes

Final answer for this problem:

$|\mathcal{E}| = \mu_0 n A \frac{dI}{dt} = (4\pi \times 10^{-7})(1500)(2.0 \times 10^{-4})(20) \approx 7.54 \times 10^{-6}\ \text{V}$