A point charge `3.2 xx 10^(-19) C " makes " 7 xx 10^(15)` revolutions per second in a circular path of radius `1 Å` . Calculate the magnetic induction at the centre of the circular path.

Answer

To solve the problem of calculating the magnetic induction at the center of a circular path due to a point charge, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Charge (q) = \(3.2 \times 10^{-19} \, \text{C}\)
- Number of revolutions per second (n) = \(7 \times 10^{15} \, \text{rev/s}\)
- Radius of the circular path (r) = \(1 \, \text{Å} = 1 \times 10^{-10} \, \text{m}\)

2. **Calculate the Total Charge Flowing per Second:**
- The total charge (Q) that crosses a point in one second can be calculated using the formula:
\[ Q = n \times q \]
- Substituting the values:
\[ Q = (7 \times 10^{15}) \times (3.2 \times 10^{-19}) = 2.24 \times 10^{-3} \, \text{C} \]

3. **Calculate the Current (I):**
- Current (I) is defined as the charge flowing per unit time. Since we are considering the charge that crosses a point in one second:
\[ I = Q = 2.24 \times 10^{-3} \, \text{A} \]

4. **Use the Formula for Magnetic Field (B) at the Center of a Circular Path:**
- The magnetic field at the center of a circular loop carrying current 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).

5. **Substituting the Values:**
- Now substituting the values of \(\mu_0\), I, and r into the formula:
\[ B = \frac{(4\pi \times 10^{-7}) \times (2.24 \times 10^{-3})}{2 \times (1 \times 10^{-10})} \]
- Simplifying this:
\[ B = \frac{4\pi \times 10^{-7} \times 2.24 \times 10^{-3}}{2 \times 10^{-10}} = \frac{4\pi \times 2.24 \times 10^{-7} \times 10^{10}}{2} \]
\[ B = 14.074 \, \text{T} \]

6. **Final Answer:**
- The magnetic induction at the center of the circular path is approximately \(14.074 \, \text{T}\).