A circular coil of radius 6 cm and 20 turns rotates about its vertical diameter with an angular speed of `40"rad"s^(-1)` in a uniform horizontal magnetic field of magnitude `2xx10^(-2)T`. If the coil form a closed loop of resistance `8omega`, then the average power loss due to joule heating is

Options

A. `2.07xx10^(-3)W`
B. `1.23xx10^(-3)W`
C. `3.14xx10^(-3)W`
D. `1.80xx10^(-3)W`

Explanation

To solve the problem step by step, we will calculate the average power loss due to Joule heating in the circular coil.

### Step 1: Calculate the Area of the Coil
The area \( A \) of a circular coil is given by the formula:
\[ A = \pi r^2 \]
where \( r \) is the radius of the coil. Given that the radius is 6 cm, we first convert it to meters:
\[ r = 6 \, \text{cm} = 0.06 \, \text{m} \]
Now, substituting the value of \( r \):
\[ A = \pi (0.06)^2 = \pi (0.0036) \approx 0.011304 \, \text{m}^2 \]

### Step 2: Calculate the Maximum EMF (E)
The maximum electromotive force (EMF) induced in the coil can be calculated using the formula:
\[ E = N A B \omega \]
where:
- \( N = 20 \) (number of turns),
- \( B = 2 \times 10^{-2} \, \text{T} \) (magnetic field strength),
- \( \omega = 40 \, \text{rad/s} \) (angular speed).

Substituting the values:
\[ E = 20 \times 0.011304 \times (2 \times 10^{-2}) \times 40 \]
Calculating this step-by-step:
\[ E = 20 \times 0.011304 \times 0.02 \times 40 \]
\[ E = 20 \times 0.011304 \times 0.8 \]
\[ E = 20 \times 0.0090432 \approx 0.180864 \, \text{V} \approx 0.18 \, \text{V} \]

### Step 3: Calculate the Current (I)
Using Ohm's law, the current \( I \) through the coil can be calculated as:
\[ I = \frac{E}{R} \]
where \( R = 8 \, \Omega \) (resistance). Substituting the values:
\[ I = \frac{0.18}{8} = 0.0225 \, \text{A} \approx 0.023 \, \text{A} \]

### Step 4: Calculate the Average Power (P)
The average power \( P \) dissipated due to Joule heating in the coil is given by:
\[ P = \frac{E I}{2} \]
Substituting the values of \( E \) and \( I \):
\[ P = \frac{0.18 \times 0.023}{2} \]
Calculating this step-by-step:
\[ P = \frac{0.00414}{2} = 0.00207 \, \text{W} = 2.07 \times 10^{-3} \, \text{W} \]

### Final Answer
The average power loss due to Joule heating is:
\[ \boxed{2.07 \times 10^{-3} \, \text{W}} \]