Q.A circular coil of radius $10\ \text{cm}$, $500$ turns and resistance $2\ \Omega$ is placed with its plane perpendicular to the horizontal component of the earth's magnetic field. It is rotated about its vertical diameter through $180^\circ$ in $0.25\ \text{s}$. Estimate the magnitudes of the emf and current induced in the coil. Horizontal component of the earth's magnetic field at the place is $3.0 \times 10^{-5}\ \text{T}$.

Concept: Electromagnetic Induction — change in magnetic flux through the coil induces an emf.

Step 1 – Initial and final flux

Area of coil: $A = \pi r^2 = \pi (0.10)^2 = 0.01\pi\ \text{m}^2$

Initial flux: $\Phi_i = N B A \cos 0^\circ = 500 \times (3.0\times10^{-5}) \times 0.01\pi = 1.5\pi \times 10^{-4}\ \text{Wb}$

After $180^\circ$ rotation, $\Phi_f = - \Phi_i$ (cosine reverses sign).

Step 2 – Change in flux

$|\Delta\Phi| = |\Phi_f - \Phi_i| = 2 \Phi_i = 3.0\pi \times 10^{-4}\ \text{Wb}$

Step 3 – Induced emf

From Faraday’s law: …

Step-by-Step Solution

Initially the plane is perpendicular to $\vec{B}$, so the normal is along $\vec{B}$ and the flux per turn is $\Phi_i=BA$. After a $180^\circ$ turn the normal reverses, so $\Phi_f=-BA$. Change in flux linkage:

Area:

Average induced emf: …

Method: Faraday’s Law of Electromagnetic Induction

This problem is solved using Faraday’s Law, which states that the induced emf in a coil is equal to the negative rate of change of magnetic flux through it.

Step-by-step solution

Step 1: Identify the change in flux

• Initial position: Plane of coil is perpendicular to the horizontal magnetic field $B_H$. → Angle between area vector $\vec{A}$ and $\vec{B}$ is $0^\circ$. → Initial flux:

• Final position: Coil rotated by $180^\circ$ about vertical diameter. → Area vector now points opposite to $\vec{B}$. → Angle = $180^\circ$, so $\cos 180^\circ = -1$ → Final flux:

Step 2: Calculate the change in flux

The magnitude of change is:

Step 3: Compute area of the coil

Radius $r = 10\ \text{cm} = 0.1\ \text{m}$

Step 4: Plug values into Faraday’s Law

$N = 500$, $B_H = 3.0 \times 10^{-5}\ \text{T}$, $\Delta t = 0.25\ \text{s}$

Step 5: Simplify

$|\mathcal{E}| = \frac{2 \times 500 \times 3.0 \times 10^{-5} \times 0.01\pi}{0.25}$ …

Here’s a breakdown of the common mistakes students make on this exact problem and how to avoid each one.

1. Forgetting to Multiply by the Number of Turns ($N$)

The Mistake:

Students often calculate the change in flux through a single turn and then forget to multiply by $N = 500$ when finding the induced emf.

Why it happens:

The formula for magnetic flux $\phi = BA\cos\theta$ is usually taught for a single loop. When a coil has $N$ turns, the total flux linkage is $N\phi$, not just $\phi$.

How to avoid:

Always write the flux linkage explicitly:

Then use Faraday’s law:

Key result:

Here, $N = 500$, $A = \pi (0.10)^2$, so the emf will be 500 times larger than for a single turn.

2. Using the Wrong Angle Change ($\Delta\theta$)

The Mistake:

Students think rotating by $180^\circ$ means the angle changes from $0^\circ$ to $180^\circ$, so they use $\Delta\theta = 180^\circ$ in a formula like $\mathcal{E} = NBA\omega \sin\theta$ incorrectly.

Why it happens:

They confuse the instantaneous emf formula (which uses $\sin\theta$) with the average emf formula (which uses $\Delta\cos\theta$).

How to avoid:

For a rotation through $180^\circ$:

• Initial angle: $\theta_i = 0^\circ$ (plane perpendicular to field → normal parallel to field)
• Final angle: $\theta_f = 180^\circ$ (normal now opposite direction)

So:

Change in $\cos\theta$:

Magnitude of change in flux linkage:

Key result:

The factor is 2, not 1 or 0.

3. Using the Wrong Area ($A$)

The Mistake:

Students use the diameter ($10\ \text{cm}$) as the radius, or forget to convert cm to m.

Why it happens:

Rushing through unit conversion.

How to avoid:

Always convert to SI units first:

• Radius $r = 10\ \text{cm} = 0.10\ \text{m}$
• Area $A = \pi r^2 = \pi (0.10)^2 = 0.01\pi\ \text{m}^2$

Key result:

$A = 3.14 \times 10^{-2}\ \text{m}^2$ (approximately).

4. Confusing Average emf with Instantaneous emf

The Mistake:

Students try to use $\mathcal{E} = NBA\omega\sin\omega t$ for this problem, which gives the instantaneous emf at a given time, not the average emf over the rotation.

Why it happens:

The problem asks for “the magnitude of the emf” — but since the rotation is at constant angular speed over a finite time, the induced emf varies. The question expects the average emf.

How to avoid:

Use the average emf formula:

Here:

Key result:

Plug in $N=500$, $B=3.0\times10^{-5}$, $A=0.01\pi$, $\Delta t=0.25$:

5. Forgetting to Calculate the Induced Current

The Mistake:

Students stop after finding the emf and don’t compute the current using Ohm’s law.

Why it happens: …