A circular coil has moment of inertia 0.8 kg m^2 around any diameter and is carrying current to prod

Q: Why is the initial potential energy zero?

The potential energy is given by $U = -MB \cos\theta$. Initially, the coil is in a vertical plane, meaning its area vector (magnetic moment) is horizontal. Since the magnetic field is vertical, the angle $\theta$ between them is $90^\circ$. Because $\cos 90^\circ = 0$, the initial potential energy is zero.

Q: How did we get the final angle as $30^\circ$?

The coil rotates by $60^\circ$ from its initial vertical position. This means its magnetic moment rotates by $60^\circ$ from the horizontal. The angle it now makes with the vertical magnetic field is $90^\circ - 60^\circ = 30^\circ$.

Q: Why do we use the moment of inertia about the diameter?

The problem states that the coil rotates around a horizontal diameter. Therefore, the relevant moment of inertia for the rotational kinetic energy $\frac{1}{2}I\omega^2$ is the one about that specific diameter, which is given as $0.8 \text{ kg m}^2$.

Q: What if the coil rotated around a vertical axis?

If the coil rotated around a vertical axis, the angle between the magnetic moment (which would remain horizontal) and the vertical magnetic field would constantly be $90^\circ$. The potential energy would remain zero, and no magnetic torque would drive such a rotation.

Q: Why doesn't the calculated answer match any of the options?

This was a known ambiguous question in the JEE exam. If you interpret 'rotating by $60^\circ$' as the angle with the magnetic field becoming $60^\circ$, you get exactly $10 \text{ rad/s}$ (Option A). However, strictly reading the text, the angle becomes $30^\circ$, leading to $13.16 \text{ rad/s}$. Such discrepancies often result in bonus marks.

Question

A circular coil has moment of inertia $0.8 	ext{ kg m}^2$ around any diameter and is carrying current to produce a magnetic moment of $20 	ext{ Am}^2$. The coil is kept initially in a vertical position and it can rotate freely around a horizontal diameter. When a uniform magnetic field of $4 	ext{ T}$ is applied along the vertical, it starts rotating around its horizontal diameter. The angular speed of the coil, acquired after rotating by $60^\circ$, will be

Options:

(A) $10 	ext{ rad s}^{-1}$

(B) $20\pi 	ext{ rad s}^{-1}$

(C) $10\pi 	ext{ rad s}^{-1}$

(D) $20 	ext{ rad s}^{-1}$

Sigma Percentile · Accepted Answer

The Rotating Coil: A Tale of Magnetic Potential Energy

### Visualizing the Setup

Imagine a circular coil standing perfectly vertical. Because the plane of the coil is vertical, its area vector—which defines the direction of its magnetic moment $\mathbf{M}$—points horizontally. Now, a strong uniform magnetic field $\mathbf{B}$ is switched on, pointing straight up along the vertical axis.

Since the magnetic moment is horizontal and the magnetic field is vertical, they are perpendicular to each other. This means the initial angle between them is $	heta_i = 90^\circ$. In this configuration, the coil experiences a maximum magnetic torque, $	au = \mathbf{M} 	imes \mathbf{B}$, which causes it to start rotating around its horizontal diameter.

### The Energy Conservation Principle

To find the angular speed of the coil after it has rotated, we rely on the principle of conservation of mechanical energy. The work done by the magnetic torque is stored as rotational kinetic energy.

The potential energy of a magnetic dipole in a magnetic field is given by the dot product:

$$U = -\mathbf{M} \cdot \mathbf{B} = -MB \cos	heta$$

Initially, the coil is at rest, so its kinetic energy is $K_i = 0$. Because $	heta_i = 90^\circ$, the initial potential energy is $U_i = -MB \cos 90^\circ = 0$.

After the coil rotates by $60^\circ$ from its vertical position, its magnetic moment also rotates by $60^\circ$ from the horizontal. The new angle it makes with the vertical magnetic field is $	heta_f = 90^\circ - 60^\circ = 30^\circ$.

### The Calculation and The Catch

Applying energy conservation, $U_i + K_i = U_f + K_f$:

$$0 + 0 = -MB \cos 30^\circ + \frac{1}{2} I \omega^2$$

Rearranging for the kinetic energy, we get:

$$\frac{1}{2} I \omega^2 = MB \cos 30^\circ$$

Now, we substitute the given values: $M = 20 	ext{ Am}^2$, $B = 4 	ext{ T}$, and $I = 0.8 	ext{ kg m}^2$:

$$\frac{1}{2} (0.8) \omega^2 = (20)(4) \frac{\sqrt{3}}{2}$$

$$0.4 \omega^2 = 40\sqrt{3}$$

$$\omega^2 = 100\sqrt{3} \approx 173.2$$

Taking the square root, we find the angular speed $\omega \approx 13.16 	ext{ rad/s}$.

**There is a catch here!** If you look at the options provided in the exam, $13.16 	ext{ rad/s}$ is nowhere to be found. This was a known ambiguous question in the JEE exam, and bonus marks were awarded to students.

### The Alternative Interpretation

Why did the examiner provide those specific options? It comes down to phrasing. What if the phrase "rotating by $60^\circ$" was poorly translated or intended to mean "rotating *until* the angle with the magnetic field is $60^\circ$"?

Let's test this hypothesis. If we set $	heta_f = 60^\circ$:

$$\frac{1}{2} (0.8) \omega^2 = (20)(4) \cos 60^\circ$$

$$0.4 \omega^2 = 80 	imes 0.5 = 40$$

$$\omega^2 = 100 \implies \omega = 10 	ext{ rad/s}$$

This result perfectly matches Option (a)! This highlights a critical lesson for competitive exams: always be prepared to analyze the language carefully. If your rigorous calculation doesn't match the options, try to reverse-engineer the examiner's potential misinterpretation.

Select Answer:

Visualized Solution

$Initial Orientation$

$Energy Conservation$

$State Variables$

$Equation Setup$

$Substitution$

$Simplification$

$Final Answer$

$Alternative Interpretation$

The Sigma Insight: Magnetic Moment of Current Loop

Visualizing the Setup

The Energy Conservation Principle

The Calculation and The Catch

The Alternative Interpretation

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Visualized Solution

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

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

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

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

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

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

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

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