A circular insulated copper wire loop is twisted to form two loops of area A and 2A as shown in the

Q: Why do we subtract the areas instead of adding them?

Because the wire is twisted into a figure-8, any current flowing through it will circulate clockwise in one loop and counter-clockwise in the other. This means their magnetic moments and area vectors oppose each other, so the effective area is the difference between the two.

Q: What does 'plane of the loops is perpendicular to the paper' mean?

Initially, the loops lie flat on the paper. As they rotate about the vertical axis, they turn. When they have rotated by 90 degrees ($\pi/2$), they are seen edge-on, meaning their plane is perpendicular to the paper.

Q: Why is the rate of change of flux maximum when the flux itself is zero?

Flux is proportional to $\cos(\omega t)$, and its derivative (rate of change) is proportional to $\sin(\omega t)$. When the loops are perpendicular to the paper, the magnetic field skims past them, so flux is zero. But at this exact moment, the flux is changing most rapidly!

Q: Why is option (c) incorrect?

The magnetic flux $\Phi$ is proportional to $\cos(\omega t)$. The induced emf is the derivative of flux, which gives a $-\sin(\omega t)$ term. Therefore, the emf is proportional to $\sin(\omega t)$, not $\cos(\omega t)$.

Q: Does the axis of rotation affect the answer?

As long as the axis of rotation is in the plane of the loops and perpendicular to the magnetic field, the flux will vary as $\cos(\omega t)$. The common diameter perfectly satisfies this condition.

Question

A circular insulated copper wire loop is twisted to form two loops of area $A$ and $2A$ as shown in the figure. At the point of crossing, the wires remain electrically insulated from each other. The entire loop lies in the plane (of the paper). A uniform magnetic field $\mathbf{B}$ points into the plane of the paper. At $t = 0$, the loop starts rotating about the common diameter as axis with a constant angular velocity $\omega$ in the magnetic field. Which of the following options is/are correct?

Options:

(A) The emf induced in the loop is proportional to the sum of the areas of the two loops.

(B) The rate of change of the flux is maximum when the plane of the loops is perpendicular to plane of the paper.

(C) The net emf induced due to both the loops is proportional to $\cos \omega t$.

(D) The amplitude of the maximum net emf induced due to both the loops is equal to the amplitude of maximum emf induced in the smaller loop alone.

Sigma Percentile · Accepted Answer

The problem of the twisted loop is a classic test of your physical intuition. It's easy to look at two loops of area $A$ and $2A$ and assume the total area is $3A$. But physics is not just about adding numbers; it's about understanding the geometry of the setup!

### Analyzing the Setup

Imagine an ant walking along this twisted wire. If it walks clockwise in the top loop, it will cross the junction and naturally walk counter-clockwise in the bottom loop. This twist is the most critical part of the problem.

Because the sense of traversal in the two loops is opposite, their area vectors must point in opposite directions. By the right-hand rule, if the top loop's area vector points out of the page, the bottom loop's area vector points into the page.

### The Master Equation

Since the area vectors oppose each other, the net area of the system is the difference between the two areas, not the sum.

$$A_{	ext{net}} = 2A - A = A$$

Now, the entire loop system is rotating with a constant angular velocity $\omega$. At any time $t$, the angle between the magnetic field $\mathbf{B}$ and the net area vector $\mathbf{A}_{	ext{net}}$ is $	heta = \omega t$.

The magnetic flux $\Phi$ passing through the system is given by the dot product:

$$\Phi = \mathbf{B} \cdot \mathbf{A}_{	ext{net}} = B A \cos(\omega t)$$

### Final Calculation

According to Faraday's Law of Induction, the induced emf $\varepsilon$ is the negative rate of change of magnetic flux. Let's differentiate our flux equation:

$$\varepsilon = -\frac{d\Phi}{dt} = -\frac{d}{dt}(B A \cos\omega t)$$

$$\varepsilon = B A \omega \sin(\omega t)$$

Now we have our master equation for the induced emf. Let's evaluate the given options!

**Option (a):** The emf is proportional to the difference of the areas ($A$), not the sum ($3A$). So, this is incorrect.

**Option (c):** The net emf is proportional to $\sin(\omega t)$, not $\cos(\omega t)$. So, this is also incorrect.

**Option (b):** The rate of change of flux is simply the magnitude of the induced emf, $\left| \frac{d\Phi}{dt} ight| = B A \omega |\sin(\omega t)|$. This value is maximum when $\sin(\omega t) = \pm 1$, which occurs at $\omega t = \pi/2$ or $3\pi/2$. At these angles, the loop has rotated by 90 degrees, meaning its plane is exactly perpendicular to the paper. Thus, this statement is **correct**.

**Option (d):** The amplitude (maximum value) of the net emf is $B A \omega$. If we consider only the smaller loop (which has an area of $A$), its maximum emf would also be $B A \omega$. They are exactly equal! Thus, this statement is also **correct**.

By carefully analyzing the geometry of the twist, we avoided the trap of adding the areas and flawlessly derived the correct physical behavior of the system!

Select Answer:

Visualized Solution

$T w i s t e d L oo p G eo m e t r y$

$N e t A r e a V ec t or$

$M a g n e t i c F l ux$

$I n d u ce d E M F$

$E v a l u a t in g O pt i o n s (a) & (c)$

$E v a l u a t in g O pt i o n (b)$

$E v a l u a t in g O pt i o n (d)$

The Sigma Insight: Faraday's Laws of Electromagnetic Induction

Analyzing the Setup

The Master Equation

Final Calculation

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