Q.A square loop of side $10\ \text{cm}$ and resistance $0.5\ \Omega$ is placed vertically in the east-west plane. A uniform magnetic field of $0.10\ \text{T}$ is set up across the plane in the north-east direction. The magnetic field is decreased to zero in $0.70\ \text{s}$ at a steady rate. Determine the magnitudes of induced emf and current during this time-interval.

Concept: Electromagnetic Induction — the induced emf is given by Faraday’s law: $\mathcal{E} = -\frac{d\Phi}{dt}$, where $\Phi = \mathbf{B} \cdot \mathbf{A}$.

Step 1: Find the initial magnetic flux.

Area of the loop: $A = (0.10\ \text{m})^2 = 0.01\ \text{m}^2$.

The field is in the north-east direction, and the loop is in the east-west plane. The normal to the loop (east-west plane) is along the north-south direction. The angle between the field (NE) and the normal (N–S) is $45^\circ$.

So, $\Phi_i = B A \cos 45^\circ = (0.10)(0.01)\left(\frac{1}{\sqrt{2}}\right) = \frac{0.001}{\sqrt{2}}\ \text{Wb}$.

Step 2: Final flux and change in flux.

Final field is zero, so $\Phi_f = 0$.

Change in flux: $|\Delta\Phi| = \frac{0.001}{\sqrt{2}}\ \text{Wb}$.

Step 3: Induced emf and current.

Time interval: $\Delta t = 0.70\ \text{s}$. …

The core idea here is electromagnetic induction: a changing magnetic flux through a loop induces an emf. The flux depends on three things — the field strength $B$, the area $A$ of the loop, and the angle between the field and the normal to the loop. In this problem, only $B$ changes, and it does so uniformly.

Let’s unpack the geometry first. The loop is vertical and lies in the east-west plane. That means its plane contains the east-west direction and the vertical direction. The normal to the loop (the direction perpendicular to its plane) therefore points north-south. The magnetic field is given as $0.10\ \text{T}$ in the north-east direction. So the field is at an angle to the normal.

We need the component of the field that actually passes through the loop — that is, the component along the normal. That’s $B \cos \theta$, where $\theta$ is the angle between the field direction and the normal.

Now let’s go step by step.

1. Find the area of the loop.

   Side length $= 10\ \text{cm} = 0.10\ \text{m}$.

   Area $A = (0.10)^2 = 1.0 \times 10^{-2}\ \text{m}^2$.

2. Find the initial magnetic flux through the loop.

   Flux $\Phi = B A \cos \theta$.

   Here $B = 0.10\ \text{T}$, $A = 1.0 \times 10^{-2}\ \text{m}^2$, $\cos 45^\circ = 1/\sqrt{2}$.

   So

3. Find the final flux.

   The field is decreased to zero, so $B_f = 0$ and therefore $\Phi_f = 0$.

4. Calculate the change in flux.

The magnitude is $|\Delta \Phi| = \frac{1.0 \times 10^{-3}}{\sqrt{2}}\ \text{Wb}$.

5. Apply Faraday’s law to find induced emf. Faraday’s law:

The time interval is $\Delta t = 0.70\ \text{s}$.

So

Compute: $0.70 \times \sqrt{2} \approx 0.70 \times 1.414 = 0.9898 \approx 0.99$.

So

Method: Faraday’s Law of Electromagnetic Induction

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

Step 1: Identify the given data

• Side of square loop, $a = 10\ \text{cm} = 0.10\ \text{m}$
• Area of loop, $A = a^2 = (0.10)^2 = 0.01\ \text{m}^2$
• Resistance, $R = 0.5\ \Omega$
• Initial magnetic field, $B_i = 0.10\ \text{T}$
• Final magnetic field, $B_f = 0\ \text{T}$
• Time interval, $\Delta t = 0.70\ \text{s}$
• Field direction: north-east (at $45^\circ$ to the east-west plane)

Step 2: Find the angle between field and area vector

The loop is in the east-west vertical plane.

The area vector is perpendicular to the loop — pointing north (or south).

The magnetic field is north-east — at $45^\circ$ to north.

So, the angle between $\vec{B}$ and area vector $\vec{A}$ is:

Step 3: Calculate initial magnetic flux

Magnetic flux:

Step 4: Calculate final flux

Since $B_f = 0$:

Step 5: Apply Faraday’s Law for induced emf

Magnitude of induced emf: …

Here are the common mistakes students make on this Electromagnetic Induction problem, along with how to avoid each.

Mistake 1: Getting the area vector direction wrong

The error:

Students often take the area vector as simply "up" or "perpendicular to the loop" without checking the orientation relative to the magnetic field. Here, the loop is in the east-west vertical plane, so its area vector is perpendicular to that plane — pointing either north or south.

How to avoid:

• Draw a clear diagram.
• For a loop in the east-west vertical plane, the normal is horizontal and points north (or south).
• The magnetic field is given as north-east, so the angle between the area vector (north) and the field (north-east) is $45^\circ$.

Key: Always find the angle $\theta$ between $\vec{B}$ and the area vector (not the plane of the loop).

Mistake 2: Using the wrong formula for flux change

The error:

Some students directly use $\text{emf} = B l v$ (motional emf) instead of Faraday’s law for a changing magnetic field.

How to avoid:

• Here, the field is decreasing uniformly — no motion, no velocity.
• Use Faraday’s law:

• For a uniform field and steady rate of change:

Mistake 3: Forgetting the $\cos\theta$ factor in flux

The error:

Students compute flux as $B A$ directly, ignoring the angle between $\vec{B}$ and the area vector.

How to avoid:

• Always write:

• Here, $\theta = 45^\circ$, so $\cos 45^\circ = \frac{1}{\sqrt{2}}$.
• The flux is not $B A$ — it’s $B A / \sqrt{2}$.

Mistake 4: Using the wrong area or units

The error:

Using side length in cm without converting to metres, or using perimeter instead of area.

How to avoid:

• Side $= 10\ \text{cm} = 0.1\ \text{m}$
• Area $A = (0.1)^2 = 0.01\ \text{m}^2$
• Always convert to SI units before plugging into formulas.

Mistake 5: Sign errors or ignoring magnitude

The error:

Students carry the negative sign from Faraday’s law into the final answer, or get confused about direction when only magnitude is asked.

How to avoid:

• The question asks for magnitudes of emf and current.
• Use:

• Ignore the negative sign — it only indicates direction (Lenz’s law).

Mistake 6: Using $\Delta B = B_f - B_i$ incorrectly

The error:

Some write $\Delta B = 0 - 0.10 = -0.10\ \text{T}$ and then get confused about sign.

How to avoid: …