A charge of `10^(-6)C` is describing a circular path of radius 1cm making 5 revolution per second. The magnetic induction field at the centre of the circle is

Options

A. `pixx10^(-10)T`
B. `pixx10^(-9)T`
C. `pi/2xx10^(-10)T`
D. `pi/2xx10^(-9) T`

Explanation

To solve the problem, we need to calculate the magnetic induction field at the center of a circular path described by a charge. Here are the steps to find the solution:

### Step-by-Step Solution

1. **Identify Given Data**:
- Charge, \( Q = 10^{-6} \, \text{C} \)
- Radius of the circular path, \( R = 1 \, \text{cm} = 0.01 \, \text{m} \)
- Frequency of revolution, \( f = 5 \, \text{rev/s} \)

2. **Convert Frequency to Angular Velocity**:
- Angular velocity \( \omega \) is given by the formula:
\[ \omega = 2\pi f \]
- Substituting the value of \( f \):
\[ \omega = 2\pi \times 5 = 10\pi \, \text{rad/s} \]

3. **Calculate the Induced Current**:
- The induced current \( I \) can be calculated using the formula:
\[ I = \frac{Q}{T} \]
- Where \( T \) is the time period, given by \( T = \frac{1}{f} \):
\[ T = \frac{1}{5} = 0.2 \, \text{s} \]
- Now substituting \( T \) into the current formula:
\[ I = Q \times f = 10^{-6} \times 5 = 5 \times 10^{-6} \, \text{A} \]

4. **Calculate the Magnetic Induction Field**:
- The magnetic induction field \( B \) at the center of a circular loop is given by:
\[ B = \frac{\mu_0 I}{2R} \]
- Where \( \mu_0 = 4\pi \times 10^{-7} \, \text{T m/A} \) (permeability of free space).
- Substituting the values:
\[ B = \frac{4\pi \times 10^{-7} \times 5 \times 10^{-6}}{2 \times 0.01} \]
- Simplifying:
\[ B = \frac{20\pi \times 10^{-13}}{0.02} = 10\pi \times 10^{-11} = \pi \times 10^{-10} \, \text{T} \]

5. **Final Answer**:
- The magnetic induction field at the center of the circle is:
\[ B = \pi \times 10^{-10} \, \text{T} \]

### Summary
The magnetic induction field at the center of the circular path is \( \pi \times 10^{-10} \, \text{T} \).