A cathode ray beam is bent in a circle of radius 2 cm by a magnetic induction `4.5 xx10^(-3)" weber//m^2` ,The velocity of electron is

Options

A. `3.43 xx10^(7)m//s`
B. `5.37 xx10^(7)m//s`
C. `1.23 xx10^(7)m//s`
D. `1.58 xx10^(7)m//s`

Explanation

To solve the problem of finding the velocity of an electron in a cathode ray beam bent in a circle by a magnetic field, we can follow these steps:

### Step-by-Step Solution:

1. **Identify Given Values:**
- Radius of the circular path (r) = 2 cm = \(2 \times 10^{-2}\) m
- Magnetic induction (B) = \(4.5 \times 10^{-3}\) Wb/m²
- Charge of the electron (q) = \(1.6 \times 10^{-19}\) C
- Mass of the electron (m) = \(9.1 \times 10^{-31}\) kg

2. **Understand the Forces Involved:**
- The centripetal force required to keep the electron moving in a circular path is provided by the magnetic force acting on it. The magnetic force (F) on a charged particle moving in a magnetic field is given by:
\[ F = qvB \]
- The centripetal force (Fc) required for circular motion is given by:
\[ F_c = \frac{mv^2}{r} \]

3. **Set the Forces Equal:**
- For the electron to move in a circle, the magnetic force must equal the centripetal force:
\[ qvB = \frac{mv^2}{r} \]

4. **Rearrange the Equation:**
- We can cancel one \(v\) from both sides (assuming \(v \neq 0\)):
\[ qB = \frac{mv}{r} \]
- Now, rearranging to solve for velocity \(v\):
\[ v = \frac{qBr}{m} \]

5. **Substituting the Values:**
- Substitute the known values into the equation:
\[ v = \frac{(1.6 \times 10^{-19} \, \text{C}) \times (4.5 \times 10^{-3} \, \text{Wb/m}^2) \times (2 \times 10^{-2} \, \text{m})}{9.1 \times 10^{-31} \, \text{kg}} \]

6. **Calculate the Velocity:**
- Performing the calculation:
\[ v = \frac{(1.6 \times 10^{-19}) \times (4.5 \times 10^{-3}) \times (2 \times 10^{-2})}{9.1 \times 10^{-31}} \]
\[ v = \frac{(1.44 \times 10^{-23})}{9.1 \times 10^{-31}} \approx 1.58 \times 10^{7} \, \text{m/s} \]

### Final Answer:
The velocity of the electron is approximately \(1.58 \times 10^{7} \, \text{m/s}\).

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