If an electron is moving in a magnetic field of `5.4 xx 10^(-4)T` on a circular path of radius 32 cm having a frequency of 2.5 MHz, then its speed will be

Options

A. `8.56xx10^(6)m s^(-1)`
B. `5.024xx10^(6)m s^(-1)`
C. `8.56xx10^(4)m s^(-1)`
D. `5.024xx10^(4)m s^(-1)`

Explanation

To find the speed of an electron moving in a magnetic field, we can use the relationship between speed, radius, and angular frequency. Here’s a step-by-step solution:

### Step 1: Understand the given values
- Magnetic field (B) = \(5.4 \times 10^{-4} \, T\)
- Radius of the circular path (r) = 32 cm = \(0.32 \, m\)
- Frequency (f) = 2.5 MHz = \(2.5 \times 10^6 \, Hz\)

### Step 2: Convert the radius to meters
Since the radius is given in centimeters, we convert it to meters:
\[ r = 32 \, cm = 32 \times 10^{-2} \, m = 0.32 \, m \]

### Step 3: Calculate the angular frequency (ω)
The angular frequency (ω) can be calculated using the formula:
\[ \omega = 2\pi f \]
Substituting the value of frequency:
\[ \omega = 2 \pi (2.5 \times 10^6) \, Hz \]
Calculating this gives:
\[ \omega \approx 2 \times 3.14 \times 2.5 \times 10^6 \approx 15.7 \times 10^6 \, rad/s \]

### Step 4: Calculate the speed (v)
The speed (v) of the electron moving in a circular path is given by the formula:
\[ v = r \omega \]
Substituting the values of r and ω:
\[ v = (0.32 \, m) \times (15.7 \times 10^6 \, rad/s) \]
Calculating this gives:
\[ v \approx 5.024 \times 10^6 \, m/s \]

### Final Answer
The speed of the electron is approximately:
\[ v \approx 5.024 \times 10^6 \, m/s \]

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