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

Options

A. `1.8 xx 10^(-4) T`
B. `5.6 xx 10^(-5) T`
C. `14.4 xx 10^(-6) T`
D. `1.3 xx 10^(-6) T`

Explanation

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

\[ r = \frac{mv}{Bq} \]

Where:
- \( r \) is the radius of the circular path (0.10 m)
- \( m \) is the mass of the electron (\( 9 \times 10^{-31} \) kg)
- \( v \) is the velocity of the electron (\( 10^6 \) m/s)
- \( B \) is the magnetic field intensity (which we need to find)
- \( q \) is the charge of the electron (\( 1.6 \times 10^{-19} \) C)

We can rearrange the formula to solve for \( B \):

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

Now, we can substitute the known values into the equation:

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

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

3. **Calculating the denominator:**
\[ \text{Denominator} = 0.10 \times 1.6 \times 10^{-19} = 0.16 \times 10^{-19} = 1.6 \times 10^{-20} \, \text{C m} \]

4. **Now substituting back into the equation for B:**
\[ B = \frac{9 \times 10^{-25}}{1.6 \times 10^{-20}} = 5.625 \times 10^{-5} \, \text{T} \]

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

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

### Final Answer:
The intensity of the magnetic field is \( 5.6 \times 10^{-5} \, \text{T} \).

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