A planar loop of wire rotates in a uniform magnetic field. Initially at t = 0, the plane of the loop

Q: Why is the angle $\theta = 0^\circ$ initially when the plane is perpendicular to the field?

The angle $\theta$ in the magnetic flux formula $\Phi = BA \cos\theta$ is the angle between the magnetic field vector and the **area vector** of the loop. The area vector is always perpendicular to the plane of the loop. So, if the plane is perpendicular to the field, the area vector is parallel to the field, making $\theta = 0^\circ$.

Q: Why do we differentiate the flux to find the emf?

According to Faraday's Law of Electromagnetic Induction, an electromotive force (emf) is induced in a circuit whenever there is a change in the magnetic flux linked with it. The magnitude of this induced emf is directly proportional to the rate of change of magnetic flux, mathematically expressed as $e = -d\Phi/dt$.

Q: Why is the maximum emf at $\omega t = \pi/2$ and not $\omega t = 0$?

At $t=0$, the flux is maximum because the loop is fully facing the magnetic field. However, the *rate of change* of flux is zero at this instant (the tangent to the cosine curve is horizontal). The flux changes most rapidly when the loop is parallel to the field ($\omega t = \pi/2$), which is when the induced emf is maximum.

Q: Does the negative sign in Faraday's Law affect the maximum magnitude?

The negative sign represents Lenz's Law, indicating the direction of the induced emf (it opposes the change in flux). When we are asked for the *magnitude* of the induced emf, we take the absolute value, so the negative sign doesn't change the times at which the maximums and minimums occur.

Q: How did we get $\omega = \pi/5$?

The angular velocity $\omega$ is related to the time period $T$ by the formula $\omega = 2\pi/T$. Since the problem states the loop completes one full rotation in $T = 10$ seconds, we substitute this to get $\omega = 2\pi/10 = \pi/5$ radians per second.

Question

A planar loop of wire rotates in a uniform magnetic field. Initially at $t =0$, the plane of the loop is perpendicular to the magnetic field. If it rotates with a period of $10	ext{ s}$ about an axis in its plane, then the magnitude of induced emf will be maximum and minimum respectively at

Options:

(A) $2.5	ext{ s}$ and $7.5	ext{ s}$

(B) $2.5	ext{ s}$ and $5.0	ext{ s}$

(C) $5.0	ext{ s}$ and $10.0	ext{ s}$

(D) $5.0	ext{ s}$ and $7.5	ext{ s}$

Sigma Percentile · Accepted Answer

Imagine you are holding a rectangular wire loop and spinning it inside a powerful, invisible magnetic field. This simple mechanical action is the beating heart of almost every power plant on Earth. But how exactly does spinning a wire create electricity? Let's break down the physics and mathematics behind this beautiful phenomenon.

### Analyzing the Setup

The problem states that at $t = 0$, the plane of the loop is perfectly perpendicular to the uniform magnetic field $\vec{B}$.

In physics, we describe the orientation of a surface using an **area vector** $\vec{A}$, which always points perpendicular (normal) to the surface itself. Because the loop's plane is perpendicular to the magnetic field, its area vector $\vec{A}$ must be perfectly parallel to the magnetic field $\vec{B}$.

This means the initial angle between them is $	heta = 0^\circ$. As the loop rotates with an angular velocity $\omega$, this angle changes continuously according to the relation $	heta = \omega t$.

### The Master Equation: Magnetic Flux

To understand the induced voltage, we first need to look at the magnetic flux $\Phi$, which is essentially a measure of how many magnetic field lines are piercing through the loop.

$$ \Phi(t) = \vec{B} \cdot \vec{A} = BA \cos(	heta) $$

Substituting our time-dependent angle, we get:

$$ \Phi(t) = BA \cos(\omega t) $$

At $t = 0$, the cosine term is $1$, meaning the flux is at its absolute maximum. The loop is "catching" as much magnetic field as geometrically possible.

### Unleashing Faraday's Law

Now, here is where the magic happens. Faraday's Law of Electromagnetic Induction tells us that nature despises a change in magnetic flux. When the flux changes, an electromotive force (EMF) is induced to fight that change.

$$ e = -\frac{d\Phi}{dt} $$

Let's differentiate our flux equation with respect to time:

$$ e = -\frac{d}{dt}(BA \cos(\omega t)) $$

Since the derivative of $\cos(\omega t)$ is $-\omega \sin(\omega t)$, the negative signs beautifully cancel out:

$$ e = BA\omega \sin(\omega t) $$

### Finding the Maximum and Minimum EMF

The question asks for the times when the **magnitude** of this induced EMF is maximum and minimum.

The magnitude $|e|$ is maximum when the absolute value of the sine function is $1$.

$$ |\sin(\omega t)| = 1 \implies \omega t = \frac{\pi}{2}, \frac{3\pi}{2}, \dots $$

We are given that the loop completes one full rotation in a time period $T = 10	ext{ s}$. This allows us to calculate the angular velocity $\omega$:

$$ \omega = \frac{2\pi}{T} = \frac{2\pi}{10} = \frac{\pi}{5}	ext{ rad/s} $$

Substituting this into our maximum condition for the first occurrence:

$$ \left(\frac{\pi}{5}ight)t = \frac{\pi}{2} $$

$$ t = \frac{5}{2} = 2.5	ext{ s} $$

So, the EMF hits its maximum magnitude at $2.5	ext{ s}$. Notice that at this exact moment, the loop has rotated by $90^\circ$, meaning its plane is parallel to the magnetic field. The flux is zero, but the *rate of change* of flux is at its absolute peak!

### Final Calculation for the Minimum

Conversely, the magnitude of the EMF is minimum (which is zero) when the sine function evaluates to zero.

$$ \sin(\omega t) = 0 \implies \omega t = \pi, 2\pi, \dots $$

For the first occurrence after $t=0$:

$$ \left(\frac{\pi}{5}ight)t = \pi $$

$$ t = 5.0	ext{ s} $$

At $5.0	ext{ s}$, the loop has rotated by $180^\circ$. It is once again perpendicular to the magnetic field, catching maximum flux, but for a brief instant, the rate of change of that flux is zero.

Therefore, the magnitude of the induced EMF is maximum at **$2.5	ext{ s}$** and minimum at **$5.0	ext{ s}$**. This elegant interplay between sine and cosine functions is the exact reason why the electricity delivered to your home is an alternating current (AC)!

Select Answer:

Visualized Solution

$I ni t ia l S e t u p$

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

$I n d u ce d E M F$

$C o n d i t i o n f or M a x im u m E M F$

$T im e f or M a x im u m E M F$

$C o n d i t i o n f or M inim u m E M F$

$T im e f or M inim u m E M F$

$T h e W a y F or w a r d$

The Sigma Insight: Faraday's Laws of Electromagnetic Induction

Analyzing the Setup

The Master Equation

Unleashing Faraday's Law

Finding the Maximum and Minimum EMF

Final Calculation for the Minimum

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