An elliptical loop having resistance $R$, of semi-major axis $a$ and semi-minor axis $b$ is placed in a magnetic field as shown in the figure. If the loop is rotated about the X-axis with angular frequency $\omega$, then the average power loss in the loop due to joule's heating is Options: (A) $\frac{\pi^2 a^2 b^2 B^2 \omega^2}{2R}$ (B) zero (C) $\frac{\pi a b B \omega}{R}$ (D) $\frac{\pi^2 a^2 b^2 B^2 \omega^2}{R}$

### The Setup: A Dance of Geometry and Magnetism Imagine an elliptical loop resting peacefully in the $xz$-plane. It has a semi-major axis $a$ and a semi-minor axis $b$. A uniform magnetic field $\vec{B}$ is pointing straight up along the $y$-axis. Now, we introduce motion. We start spinning this loop around the $x$-axis with a constant angular frequency $\omega$. As the loop rotates, the amount of magnetic field passing through its surface constantly changes. This changing magnetic flux is the engine that will drive an induced current through the loop. ### The Engine of Induction: Faraday's Law To find the induced EMF, we first need to write down the magnetic flux $\Phi$. Flux is the dot product of the magnetic field and the area vector: $$\Phi = \vec{B} \cdot \vec{A} = BA \cos(\theta)$$ Because the loop is rotating at a constant rate $\omega$, the angle $\theta$ between the area vector and the magnetic field is simply $\omega t$. Now, we apply **Faraday's Law of Induction**. The induced EMF is the negative rate of change of magnetic flux: $$e = -\frac{d\Phi}{dt} = -\frac{d}{dt}(BA \cos(\omega t))$$ Differentiating this expression with respect to time gives us: $$e = BA\omega \sin(\omega t)$$ The peak value of this alternating EMF, $e_{max}$, is simply the amplitude of the sine wave: $$e_{max} = BA\omega$$ ### The Geometry of the Loop What is the area of our loop? Since it is an ellipse with semi-major axis $a$ and semi-minor axis $b$, its area is given by the geometric formula: $$A = \pi ab$$ Let's substitute this geometric fact back into our peak EMF equation. We get: $$e_{max} = B(\pi ab)\omega$$ ### Power Dissipation: The Cost of Resistance The loop has a resistance $R$, so the induced alternating current will dissipate energy as heat (Joule heating). The average power lost over a full cycle in an AC circuit is the square of the RMS voltage divided by the resistance: $$P_{avg} = \frac{e_{rms}^2}{R}$$ Remember, for a sinusoidal wave, the RMS value is the peak value divided by $\sqrt{2}$. Squaring this gives us the average power in terms of the peak EMF: $$P_{avg} = \frac{e_{max}^2}{2R}$$ ### The Grand Finale We are at the final step. Let's substitute our expression for $e_{max}$ into the average power formula. We need to square the entire term $\pi ab B \omega$: $$P_{avg} = \frac{(\pi ab B \omega)^2}{2R}$$ Carefully squaring each variable, we arrive at our final expression: $$P_{avg} = \frac{\pi^2 a^2 b^2 B^2 \omega^2}{2R}$$ This perfectly matches option (a). Notice how the power depends on the square of the magnetic field and the square of the frequency. It is a beautiful result combining geometry and electromagnetism!

JEE Main 2020

LEVELJEE Advanced

Animated Solution for Physics - Electromagnetic Induction: An elliptical loop having resistance $R$ , of semi-major axis $a$ and semi-minor axis $b$ is placed in a magnetic field as shown in the figure. If the loop is rotated about the X-axis with angular frequency $ω$ , then the average power loss in the loop due to joule's heating is

Q: Why do we use RMS voltage for average power instead of maximum voltage?

Because power depends on the square of the voltage ($P = V^2/R$). The average of $\sin^2(\omega t)$ over a full cycle is $1/2$, which naturally leads to the RMS value $V_{rms} = V_{max}/\sqrt{2}$.

Q: Why is the average power not zero if the EMF alternates?

Power is given by $V^2/R$. Since the voltage is squared, the power is always positive. This means energy is continuously dissipated as heat, regardless of the direction of the current.

Q: What if the loop was rotated about the y-axis instead?

If rotated about the $y$-axis, the area vector would remain parallel to the $y$-axis. The angle with the magnetic field wouldn't change, resulting in zero induced EMF and zero power loss.

Select Answer:

Visualized Solution

$Visualizing the Setup$

An elliptical loop is placed in the $x z$ -plane.
Semi-major axis $= a$ , Semi-minor axis $= b$ .
Uniform magnetic field $B$ is along the $y$ -axis.
The loop rotates about the $x$ -axis with angular frequency $ω$ .

$Magnetic Flux$

Magnetic flux through the loop is given by:
$Φ = B \cdot A = B A cos (θ)$
Where $θ$ is the angle between the area vector $A$ and magnetic field $B$ .
Since the loop rotates with angular frequency $ω$ , $θ = ω t$ .

$Faraday’s Law of Induction$

According to Faraday's Law, induced EMF is:
$e = - \frac{d Φ}{d t}$
$e = - \frac{d}{d t} (B A cos (ω t))$
$e = B A ω sin (ω t)$
Maximum induced EMF: $e_{ma x} = B A ω$

$Area of the Ellipse$

The area of an ellipse with semi-axes $a$ and $b$ is:
$A = π ab$
Substituting this into the peak EMF equation:
$e_{ma x} = B (π ab) ω$

$Average Power Dissipation$

The average power dissipated as Joule heating in an AC circuit is:
$P_{a v g} = \frac{e _{r m s}^{2}}{R}$
Since $e_{r m s} = \frac{e _{ma x}}{2}$ , we can write:
$P_{a v g} = \frac{e _{ma x}^{2}}{2 R}$

$Final Calculation$

Substitute $e_{ma x} = π ab B ω$ into the power equation:
$P_{a v g} = \frac{( π ab B ω ) ^{2}}{2 R}$
$P_{a v g} = \frac{π ^{2} a ^{2} b ^{2} B ^{2} ω ^{2}}{2 R}$

$Conclusion$

The average power loss in the loop due to Joule's heating is:
$P_{a v g} = \frac{π ^{2} a ^{2} b ^{2} B ^{2} ω ^{2}}{2 R}$
This matches option (a).

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The Sigma Insight: Faraday's Laws of Electromagnetic Induction

Solution Diagram

The Setup

A Dance of Geometry and Magnetism

Imagine an elliptical loop resting peacefully in the

x z

-plane. It has a semi-major axis

a

and a semi-minor axis

b

. A uniform magnetic field

B

is pointing straight up along the

y

-axis.

Now, we introduce motion. We start spinning this loop around the

x

-axis with a constant angular frequency

ω

. As the loop rotates, the amount of magnetic field passing through its surface constantly changes. This changing magnetic flux is the engine that will drive an induced current through the loop.

The Engine of Induction

Faraday's Law

To find the induced EMF, we first need to write down the magnetic flux

Φ

. Flux is the dot product of the magnetic field and the area vector:

Φ = B \cdot A = B A cos (θ)

Because the loop is rotating at a constant rate

ω

, the angle

θ

between the area vector and the magnetic field is simply

ω t

Now, we apply Faraday's Law of Induction. The induced EMF is the negative rate of change of magnetic flux:

e = - \frac{d Φ}{d t} = - \frac{d}{d t} (B A cos (ω t))

Differentiating this expression with respect to time gives us:

e = B A ω sin (ω t)

The peak value of this alternating EMF,

e_{ma x}

, is simply the amplitude of the sine wave:

e_{ma x} = B A ω

The Geometry of the Loop

What is the area of our loop? Since it is an ellipse with semi-major axis

a

and semi-minor axis

b

, its area is given by the geometric formula:

A = π ab

Let's substitute this geometric fact back into our peak EMF equation. We get:

e_{ma x} = B (π ab) ω

Power Dissipation

The Cost of Resistance

The loop has a resistance

R

, so the induced alternating current will dissipate energy as heat (Joule heating). The average power lost over a full cycle in an AC circuit is the square of the RMS voltage divided by the resistance:

P_{a v g} = \frac{e _{r m s}^{2}}{R}

Remember, for a sinusoidal wave, the RMS value is the peak value divided by

2

. Squaring this gives us the average power in terms of the peak EMF:

P_{a v g} = \frac{e _{ma x}^{2}}{2 R}

The Grand Finale

We are at the final step. Let's substitute our expression for

e_{ma x}

into the average power formula. We need to square the entire term

π ab B ω

P_{a v g} = \frac{( π ab B ω ) ^{2}}{2 R}

Carefully squaring each variable, we arrive at our final expression:

P_{a v g} = \frac{π ^{2} a ^{2} b ^{2} B ^{2} ω ^{2}}{2 R}

This perfectly matches option (a). Notice how the power depends on the square of the magnetic field and the square of the frequency. It is a beautiful result combining geometry and electromagnetism!

Common Traps & FAQs

Why do we use RMS voltage for average power instead of maximum voltage?

Because power depends on the square of the voltage (

P = V^{2} / R

). The average of

sin^{2} (ω t)

over a full cycle is

1/2

, which naturally leads to the RMS value

V_{r m s} = V_{ma x} / 2

What is the area vector of the loop initially?

The area vector is always perpendicular to the plane of the loop. If the loop is initially in the

x z

-plane, its area vector points along the

y

-axis, parallel to the magnetic field.

Does the shape of the loop matter for the EMF?

Only the total area matters for the magnitude of the magnetic flux. The elliptical shape simply dictates that we use the formula

A = π ab

Why is the average power not zero if the EMF alternates?

Power is given by

V^{2} / R

. Since the voltage is squared, the power is always positive. This means energy is continuously dissipated as heat, regardless of the direction of the current.

What if the loop was rotated about the y-axis instead?

If rotated about the

y

-axis, the area vector would remain parallel to the

y

-axis. The angle with the magnetic field wouldn't change, resulting in zero induced EMF and zero power loss.

Select Answer:

Visualized Solution

$Visualizing the Setup$

$Magnetic Flux$

$Faraday’s Law of Induction$

$Area of the Ellipse$

$Average Power Dissipation$

$Final Calculation$

$Conclusion$

The Sigma Insight: Faraday's Laws of Electromagnetic Induction

The Setup

The Engine of Induction

The Geometry of the Loop

Power Dissipation

The Grand Finale

Similar Questions

In a uniform magnetic field of induction $B$ , a wire in the form of semi-circle of radius $r$ rotates about the diameter of the circle with angular frequency $ω$ . If the total resistance of the circuit is $R$ , the mean power generated per period of rotation is

List-I

List-II

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 $10 s$ about an axis in its plane, then the magnitude of induced emf will be maximum and minimum respectively at

Comprehension Passage

The magnitude of the induced electric field in the orbit at any instant of time during the time interval of the magnetic field change is

The change in the magnetic dipole moment associated with the orbit, at the end of the time interval of the magnetic field change, is

Select Answer:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Visualizing the Setup

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Magnetic Flux

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Faraday’s Law of Induction

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Area of the Ellipse

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Average Power Dissipation

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Final Calculation

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Conclusion

The Sigma Insight: Faraday's Laws of Electromagnetic Induction

The Setup

The Engine of Induction

The Geometry of the Loop

Power Dissipation

The Grand Finale

Similar Questions

List-I

List-II

Comprehension Passage

The magnitude of the induced electric field in the orbit at any instant of time during the time interval of the magnetic field change is

The change in the magnetic dipole moment associated with the orbit, at the end of the time interval of the magnetic field change, is

$Visualizing the Setup$

$Magnetic Flux$

$Faraday’s Law of Induction$

$Area of the Ellipse$

$Average Power Dissipation$

$Final Calculation$

$Conclusion$