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: Understand the Motion of the Electron
An electron moving perpendicular to a magnetic field will experience a magnetic force that acts as a centripetal force, causing it to move in a circular path.

### Step 2: Use the Formula for Time Period
The time period \( T \) for an electron moving in a circular orbit in a magnetic field can be calculated using the formula:
\[ T = \frac{2\pi m}{qB} \]
where:
- \( m \) is the mass of the electron,
- \( q \) is the charge of the electron,
- \( B \) is the magnetic field strength.

### Step 3: Substitute the Known Values
We know:
- Mass of the electron \( m = 9.1 \times 10^{-31} \) kg,
- Charge of the electron \( q = 1.6 \times 10^{-19} \) C,
- Magnetic field \( B = 3.534 \times 10^{-5} \) T.

Substituting these values into the formula:
\[ T = \frac{2\pi (9.1 \times 10^{-31})}{(1.6 \times 10^{-19})(3.534 \times 10^{-5})} \]

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

### Step 5: Calculate the Time Period
Now substitute this back into the equation for \( T \):
\[ T = \frac{2\pi (9.1 \times 10^{-31})}{5.6544 \times 10^{-24}} \]
Calculating \( 2\pi \):
\[ 2\pi \approx 6.2832 \]
Now calculate \( T \):
\[ T = \frac{6.2832 \times 9.1 \times 10^{-31}}{5.6544 \times 10^{-24}} \approx \frac{5.72 \times 10^{-30}}{5.6544 \times 10^{-24}} \approx 1.01 \times 10^{-6} \text{ seconds} \]

### Step 6: Convert to Microseconds
Since \( 1 \text{ microsecond} = 10^{-6} \text{ seconds} \), we can express the time period as:
\[ T \approx 1 \text{ microsecond} \]

### Final Answer
Thus, the time taken by the electron to complete a circular orbit is approximately \( 1 \text{ microsecond} \).