An electron (mass `=9.0xx10^(-31)` kg and charge `=1.6xx10^(-19)` coulomb) is moving in a circular orbit in a magnetic field of `1.0xx10^(-4)"weber"//m^(2)`. Its perido of revolution is

Options

A. `3.5xx10^(-7)` second
B. `7.0xx10^(-7)` second
C. `1.05xx10^(-6)` second
D. `2.1xx10^(-6)` second

Answer

To find the period of revolution of an electron moving in a magnetic field, we can use the formula for the time period \( T \) of a charged particle moving in a circular path in a magnetic field. The formula is given by:

\[ T = \frac{2\pi m}{qB} \]

Where:
- \( T \) is the period of revolution,
- \( m \) is the mass of the electron,
- \( q \) is the charge of the electron,
- \( B \) is the magnetic field strength.

### Step 1: Identify the given values
- Mass of the electron, \( m = 9.0 \times 10^{-31} \) kg
- Charge of the electron, \( q = 1.6 \times 10^{-19} \) C
- Magnetic field strength, \( B = 1.0 \times 10^{-4} \) Wb/m²

### Step 2: Substitute the values into the formula
Substituting the values into the formula for the period of revolution:

\[ T = \frac{2\pi (9.0 \times 10^{-31})}{(1.6 \times 10^{-19})(1.0 \times 10^{-4})} \]

### Step 3: Calculate the denominator
First, calculate the denominator:

\[ (1.6 \times 10^{-19})(1.0 \times 10^{-4}) = 1.6 \times 10^{-23} \]

### Step 4: Calculate the numerator
Now calculate the numerator:

\[ 2\pi (9.0 \times 10^{-31}) \approx 2 \times 3.14 \times 9.0 \times 10^{-31} \approx 5.654 \times 10^{-30} \]

### Step 5: Calculate the period \( T \)
Now, divide the numerator by the denominator:

\[ T = \frac{5.654 \times 10^{-30}}{1.6 \times 10^{-23}} \approx 3.53 \times 10^{-7} \text{ seconds} \]

### Step 6: Round the answer
Rounding to two significant figures, we get:

\[ T \approx 3.5 \times 10^{-7} \text{ seconds} \]

Thus, the period of revolution of the electron is approximately \( 3.5 \times 10^{-7} \) seconds.

### Final Answer
The period of revolution is \( 3.5 \times 10^{-7} \) seconds.

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