An electron moves with a uniform speed of `1.8 xx 10^(6) m s^(-1)` in a circular orbit of radius `0.8 Å`. Calculate the magnetic induction at the centre of orbit.

Answer

To solve the problem of calculating the magnetic induction at the center of the orbit of an electron moving in a circular path, we will follow these steps:

### Step 1: Understand the given values
- Speed of the electron, \( v = 1.8 \times 10^6 \, \text{m/s} \)
- Radius of the orbit, \( r = 0.8 \, \text{Å} = 0.8 \times 10^{-10} \, \text{m} \)

### Step 2: Calculate the circumference of the circular path
The circumference \( C \) of the circular path is given by the formula:
\[ C = 2\pi r \]
Substituting the value of \( r \):
\[ C = 2\pi (0.8 \times 10^{-10}) \approx 5.0265 \times 10^{-10} \, \text{m} \]

### Step 3: Calculate the time period \( T \) for one complete revolution
The time period \( T \) can be calculated using the formula:
\[ T = \frac{C}{v} \]
Substituting the values:
\[ T = \frac{5.0265 \times 10^{-10}}{1.8 \times 10^6} \approx 2.79 \times 10^{-16} \, \text{s} \]

### Step 4: Calculate the current \( I \)
The current \( I \) is defined as the charge flowing per unit time. Since one electron has a charge of \( e = 1.6 \times 10^{-19} \, \text{C} \), the current due to one electron completing one revolution is:
\[ I = \frac{e}{T} \]
Substituting the values:
\[ I = \frac{1.6 \times 10^{-19}}{2.79 \times 10^{-16}} \approx 5.73 \times 10^{-4} \, \text{A} \]

### Step 5: Calculate the magnetic induction \( B \) at the center of the orbit
The magnetic field \( B \) at the center of a circular loop carrying current \( I \) is given by the formula:
\[ B = \frac{\mu_0 I}{2r} \]
Where \( \mu_0 = 4\pi \times 10^{-7} \, \text{T m/A} \). Substituting the values:
\[ B = \frac{4\pi \times 10^{-7} \times 5.73 \times 10^{-4}}{2 \times (0.8 \times 10^{-10})} \]
Calculating this gives:
\[ B \approx 4.5 \, \text{T} \]

### Final Answer
The magnetic induction at the center of the orbit is approximately \( 4.5 \, \text{T} \).

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