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 reasoning presented in the video transcript.

### Step 1: Determine the velocity of the electron
Since the electron experiences no deflection in the presence of both electric and magnetic fields, we can equate the electric force and the magnetic force acting on the electron.

The electric force \( F_E \) is given by:
\[ F_E = qE \]
where \( q \) is the charge of the electron and \( E \) is the electric field strength.

The magnetic force \( F_B \) is given by:
\[ F_B = qvB \]
where \( v \) is the velocity of the electron and \( B \) is the magnetic field strength.

Since \( F_E = F_B \), we can set these equations equal to each other:
\[ qE = qvB \]

We can cancel \( q \) from both sides (since \( q \neq 0 \)):
\[ E = vB \]

Now, we can solve for the velocity \( v \):
\[ v = \frac{E}{B} \]

### Step 2: Substitute the values for \( E \) and \( B \)
Given:
- \( E = 3.2 \times 10^5 \, \text{V/m} \)
- \( B = 2.0 \times 10^{-3} \, \text{Wb/m}^2 \)

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

### Step 3: Calculate the radius of the orbit when 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 centripetal force required for circular motion is provided by the magnetic force.

The centripetal force is given by:
\[ F_c = \frac{mv^2}{r} \]
where \( m \) is the mass of the electron and \( r \) is the radius of the orbit.

Setting the centripetal force equal to the magnetic force:
\[ \frac{mv^2}{r} = qvB \]

We can rearrange this to solve for \( r \):
\[ r = \frac{mv}{qB} \]

### Step 4: Substitute the values for \( m \), \( v \), \( q \), and \( B \)
Given:
- \( m = 9.1 \times 10^{-31} \, \text{kg} \)
- \( q = 1.6 \times 10^{-19} \, \text{C} \)
- \( B = 2.0 \times 10^{-3} \, \text{Wb/m}^2 \)
- \( v = 1.6 \times 10^8 \, \text{m/s} \)

Substituting these values into the equation for \( r \):
\[ r = \frac{(9.1 \times 10^{-31})(1.6 \times 10^8)}{(1.6 \times 10^{-19})(2.0 \times 10^{-3})} \]

### Step 5: Calculate the radius
Calculating the numerator:
\[ 9.1 \times 10^{-31} \times 1.6 \times 10^8 = 1.456 \times 10^{-22} \]

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

Now substituting these into the equation for \( r \):
\[ r = \frac{1.456 \times 10^{-22}}{3.2 \times 10^{-22}} = 0.455 \, \text{m} \]

### Final Answer
Thus, the radius of the orbit when the electric field is removed is approximately:
\[ r \approx 0.45 \, \text{m} \]