An electron `(q = 1.6 × 10^(-19) C)` is moving at right angles to a uniform magnetic field of `3.534 xx10^(-5)` T The time taken by the electron to complete a circular orbit is

Options

A. `2mus`
B. `4mus`
C. `3mus`
D. `1mus`

Answer

To solve the problem of finding the time taken by an electron to complete a circular orbit in a magnetic field, we can follow these steps:

### Step 1: Identify the Given Values
- Charge of the electron, \( q = 1.6 \times 10^{-19} \, \text{C} \)
- Magnetic field strength, \( B = 3.534 \times 10^{-5} \, \text{T} \)
- Mass of the electron, \( m = 9.1 \times 10^{-31} \, \text{kg} \)

### Step 2: Use the Formula for Time Period
The time period \( T \) for an electron moving in a circular path in a magnetic field is given by the formula:
\[ T = \frac{2\pi m}{qB} \]

### Step 3: Substitute the Values into the Formula
Now, we substitute the known values into the formula:
\[ T = \frac{2\pi (9.1 \times 10^{-31} \, \text{kg})}{(1.6 \times 10^{-19} \, \text{C})(3.534 \times 10^{-5} \, \text{T})} \]

### Step 4: Calculate the Denominator
First, calculate the denominator:
\[ (1.6 \times 10^{-19} \, \text{C})(3.534 \times 10^{-5} \, \text{T}) = 5.6544 \times 10^{-24} \, \text{C} \cdot \text{T} \]

### Step 5: Calculate the Time Period
Now, substitute the denominator back into the time period formula:
\[ T = \frac{2\pi (9.1 \times 10^{-31})}{5.6544 \times 10^{-24}} \]

Calculating this gives:
\[ T \approx \frac{2 \times 3.14 \times 9.1 \times 10^{-31}}{5.6544 \times 10^{-24}} \approx \frac{5.65 \times 10^{-30}}{5.6544 \times 10^{-24}} \approx 1 \times 10^{-6} \, \text{s} \]

### Final Answer
Thus, the time taken by the electron to complete a circular orbit is approximately:
\[ T \approx 1 \, \mu s \]