An electron (mass `=9.1xx10^(-31)kg`, charge `=1.6xx10^(-19)C`) experiences no deflection if subjected to an electric field of `3.2x10^(5)V/m`, and a magnetic fields of `2.0xx10^(-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

Answer

To solve the problem step by step, we will follow the logic presented in the video transcript and apply the relevant physics concepts.

### Step 1: Determine the velocity of the electron

Given that the electron experiences no deflection in the presence of both electric and magnetic fields, we can use the condition for a velocity selector:

\[ E = vB \]

Where:
- \( E \) is the electric field strength,
- \( v \) is the velocity of the electron,
- \( B \) is the magnetic field strength.

We can rearrange this formula to find the velocity \( v \):

\[ v = \frac{E}{B} \]

Substituting the given values:

\[ E = 3.2 \times 10^5 \, \text{V/m} \]
\[ B = 2.0 \times 10^{-3} \, \text{Wb/m}^2 \]

Calculating \( v \):

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

### Step 2: Calculate the radius of the electron's orbit after the electric field is removed

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:

\[ R = \frac{mv}{qB} \]

Where:
- \( m \) is the mass of the electron,
- \( q \) is the charge of the electron.

Substituting the known values:

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

Now substituting these values into the formula for radius:

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

Calculating the numerator:

\[ 9.1 \times 10^{-31} \times 1.6 \times 10^8 = 1.456 \times 10^{-22} \, \text{kg m/s} \]

Calculating the denominator:

\[ (1.6 \times 10^{-19}) \times (2.0 \times 10^{-3}) = 3.2 \times 10^{-22} \, \text{C m/s} \]

Now substituting back into the radius formula:

\[ R = \frac{1.456 \times 10^{-22}}{3.2 \times 10^{-22}} = 0.455 \, \text{m} \]

Rounding to two decimal places, we find:

\[ R \approx 0.45 \, \text{m} \]

### Final Answer:
The radius of the electron's orbit after the electric field is removed is approximately \( 0.45 \, \text{m} \).

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