An electron (mass = `9.1 xx 10^(-31)` kg , charge = `1.6 xx 10^(-19) C)` experiences no deflection if subjected to an electric field of `3.2 xx 10^(5)` V/m, and a magnetic field of `2.0 xx 10^(-3)`Wb/`m^(2)` Both the fields are normal to the path of electron and to each other. If the electric field is removed, then the electron will revolve in an orbit of radius

Options

A. 45 m
B. 4.5 m
C. 0.45 m
D. 0.045 m

Explanation

To solve the problem step by step, we will follow these steps:

### Step 1: Understand the Given Information
We have an electron with:
- Mass \( m = 9.1 \times 10^{-31} \, \text{kg} \)
- Charge \( q = 1.6 \times 10^{-19} \, \text{C} \)
- Electric field \( E = 3.2 \times 10^{5} \, \text{V/m} \)
- Magnetic field \( B = 2.0 \times 10^{-3} \, \text{Wb/m}^2 \)

The electron experiences no deflection when both fields are applied, indicating that the forces due to the electric and magnetic fields are equal in magnitude.

### Step 2: Calculate the Speed of the Electron
Since the forces are balanced, we can use the relationship:
\[ F_E = F_B \]
Where:
- \( F_E = qE \) (Electric force)
- \( F_B = qvB \) (Magnetic force)

Setting them equal gives:
\[ qE = qvB \]
We can cancel \( q \) (since it is non-zero):
\[ E = vB \]
From this, we can solve for the speed \( v \):
\[ v = \frac{E}{B} \]

Substituting the values:
\[ v = \frac{3.2 \times 10^{5}}{2.0 \times 10^{-3}} = 1.6 \times 10^{8} \, \text{m/s} \]

### Step 3: Calculate the Radius of the Electron's Orbit
When the electric field is removed, the electron will move in a circular path due to the magnetic field. The radius \( R \) of the circular path can be calculated using the formula:
\[ R = \frac{mv}{qB} \]

Substituting the values:
- Mass \( m = 9.1 \times 10^{-31} \, \text{kg} \)
- Speed \( v = 1.6 \times 10^{8} \, \text{m/s} \)
- Charge \( q = 1.6 \times 10^{-19} \, \text{C} \)
- Magnetic field \( B = 2.0 \times 10^{-3} \, \text{Wb/m}^2 \)

Calculating:
\[ R = \frac{(9.1 \times 10^{-31}) \times (1.6 \times 10^{8})}{(1.6 \times 10^{-19}) \times (2.0 \times 10^{-3})} \]

Calculating the numerator:
\[ 9.1 \times 10^{-31} \times 1.6 \times 10^{8} = 1.456 \times 10^{-22} \]

Calculating the denominator:
\[ 1.6 \times 10^{-19} \times 2.0 \times 10^{-3} = 3.2 \times 10^{-22} \]

Now, substituting these values into the radius formula:
\[ R = \frac{1.456 \times 10^{-22}}{3.2 \times 10^{-22}} = 0.455 \, \text{m} \approx 0.45 \, \text{m} \]

### Step 4: Conclusion
The radius of the electron's orbit when the electric field is removed is approximately \( R = 0.45 \, \text{m} \).