An electron has mass `9xx10^(-31)kg` and change `1.6xx10^(-19)C` is moving with a velocity of `10^(6)m//s`, enters a region where magnetic field exists. If it describes a circle of radius `0.10m`, the intensity of magnetic field must be

Options

A. `1.8xx10^(-4)T`
B. `5.6xx10^(-5)T`
C. `14.4xx10^(-5)T`
D. `1.3xx10^(-6)T`

Answer

To find the intensity of the magnetic field (B) required for an electron to move in a circular path of radius (R), we can use the formula that relates the magnetic field, charge, velocity, mass, and radius of the circular path:

\[ B = \frac{mv}{qR} \]

Where:
- \( m \) is the mass of the electron = \( 9 \times 10^{-31} \, \text{kg} \)
- \( q \) is the charge of the electron = \( 1.6 \times 10^{-19} \, \text{C} \)
- \( v \) is the velocity of the electron = \( 10^6 \, \text{m/s} \)
- \( R \) is the radius of the circular path = \( 0.10 \, \text{m} \)

Now, substituting the values into the formula:

1. **Substitute the values into the equation:**
\[ B = \frac{(9 \times 10^{-31} \, \text{kg}) \times (10^6 \, \text{m/s})}{(1.6 \times 10^{-19} \, \text{C}) \times (0.10 \, \text{m})} \]

2. **Calculate the numerator:**
\[ \text{Numerator} = 9 \times 10^{-31} \times 10^6 = 9 \times 10^{-25} \, \text{kg m/s} \]

3. **Calculate the denominator:**
\[ \text{Denominator} = (1.6 \times 10^{-19}) \times (0.10) = 1.6 \times 10^{-20} \, \text{C m} \]

4. **Now, divide the numerator by the denominator:**
\[ B = \frac{9 \times 10^{-25}}{1.6 \times 10^{-20}} = 5.625 \times 10^{-5} \, \text{T} \]

5. **Round the answer:**
\[ B \approx 5.6 \times 10^{-5} \, \text{T} \]

Thus, the intensity of the magnetic field must be approximately \( 5.6 \times 10^{-5} \, \text{T} \).