A circular ring of radius a is made from a wire having resistance `(lambda)` per unit length. The ring is mounted on a car such that ring remains vertical. The care moves along a horizontal circle of radius R and completes n revolutions per minute. The horizontal component of earth's magnetic field be H,
Calculate average rate at which heat is produced in the ring

Options

A. `(pi^(3)a^(3)H^(2)n^(2))/(1800 lambda)J s^(-1)`
B. `(pi^(3)a^(3)H^(2)n^(2))/(3600 lambda)J s^(-1)`
C. `(pi^(3)a^(3)H^(2)n^(2))/(1200 lambda)J s^(-1)`
D. None of these

Explanation

To solve the problem step by step, we will calculate the average rate at which heat is produced in the circular ring mounted on a car moving in a horizontal circle.

### Step 1: Determine the angular velocity of the car
The car completes \( n \) revolutions per minute. To convert this to radians per second, we use the formula:
\[ \omega = \frac{2\pi n}{60} \quad \text{(in radians per second)} \]

### Step 2: Calculate the area of the circular ring
The area \( A \) of the circular ring can be calculated using the formula for the area of a circle:
\[ A = \pi a^2 \]
where \( a \) is the radius of the ring.

### Step 3: Determine the magnetic flux through the ring
The magnetic flux \( \Phi \) through the ring is given by:
\[ \Phi = B \cdot A \]
where \( B \) is the magnetic field. Since the horizontal component of the Earth's magnetic field is \( H \) and the angle \( \theta \) changes with time, we can express the flux as:
\[ \Phi = H \cdot A \cdot \cos(\theta) \]
Substituting for \( A \):
\[ \Phi = H \cdot \pi a^2 \cdot \cos\left(\frac{n \pi t}{30}\right) \]

### Step 4: Calculate the induced EMF in the ring
The induced EMF \( \mathcal{E} \) in the ring can be found using Faraday's law of electromagnetic induction:
\[ \mathcal{E} = -\frac{d\Phi}{dt} \]
Differentiating \( \Phi \) with respect to time \( t \):
\[ \mathcal{E} = -\frac{d}{dt}\left(H \pi a^2 \cos\left(\frac{n \pi t}{30}\right)\right) = H \pi a^2 \cdot \frac{n \pi}{30} \sin\left(\frac{n \pi t}{30}\right) \]

### Step 5: Calculate the resistance of the ring
The resistance \( R \) of the ring can be calculated using the resistance per unit length \( \lambda \):
\[ R = \lambda \cdot C \]
where \( C \) is the circumference of the ring:
\[ C = 2\pi a \]
Thus,
\[ R = \lambda \cdot 2\pi a \]

### Step 6: Calculate the instantaneous power
The instantaneous power \( P \) dissipated in the ring can be calculated using:
\[ P = \frac{\mathcal{E}^2}{R} \]
Substituting for \( \mathcal{E} \) and \( R \):
\[ P = \frac{\left(H \pi a^2 \cdot \frac{n \pi}{30} \sin\left(\frac{n \pi t}{30}\right)\right)^2}{\lambda \cdot 2\pi a} \]
This simplifies to:
\[ P = \frac{H^2 \pi^4 a^4 n^2}{900 \lambda} \sin^2\left(\frac{n \pi t}{30}\right) \]

### Step 7: Calculate the average power
To find the average power \( P_{\text{avg}} \) over one complete cycle, we integrate \( P \) over one period \( T \) and divide by \( T \):
\[ P_{\text{avg}} = \frac{1}{T} \int_0^T P \, dt \]
The average value of \( \sin^2(x) \) over one period is \( \frac{1}{2} \):
\[ P_{\text{avg}} = \frac{H^2 \pi^4 a^4 n^2}{900 \lambda} \cdot \frac{T}{2} \]
Since \( T = \frac{60}{n} \):
\[ P_{\text{avg}} = \frac{H^2 \pi^4 a^4 n^2}{900 \lambda} \cdot \frac{30}{n} = \frac{H^2 \pi^4 a^4 n}{300 \lambda} \]

### Final Answer
The average rate at which heat is produced in the ring is:
\[ P_{\text{avg}} = \frac{H^2 \pi^4 a^4 n}{300 \lambda} \]