A beam of electrons is under the effect of potential difference of `1.36xx10^(4)` volt applied across two parallel plates `4 cm` apart and a magnetic field of `0.002 Wbm^(-2)` at right angles to each other. If these two field produces no deflection in the beam. find the velocity of electrons. What will be radius of the orbit in which the beam will move, if the electric field is made zero ? Given, Mass of electron `=9.1xx10^(-31)kg`.

Answer

To solve the problem step by step, we will first find the velocity of the electrons and then calculate the radius of the orbit when the electric field is made zero.

### Step 1: Calculate the Electric Field (E)
The electric field (E) between two parallel plates is given by the formula:

\[ E = \frac{V}{d} \]

where:
- \( V = 1.36 \times 10^4 \, \text{volts} \) (potential difference)
- \( d = 4 \, \text{cm} = 4 \times 10^{-2} \, \text{m} \) (distance between the plates)

Substituting the values:

\[ E = \frac{1.36 \times 10^4}{4 \times 10^{-2}} = \frac{1.36 \times 10^4}{0.04} = 3.4 \times 10^5 \, \text{V/m} \]

### Step 2: Use the Condition of No Deflection
Since the beam of electrons experiences no deflection, the electric force and magnetic force must be equal in magnitude:

\[ F_E = F_B \]

The electric force (\( F_E \)) is given by:

\[ F_E = qE \]

The magnetic force (\( F_B \)) is given by:

\[ F_B = qvB \]

Setting these equal gives:

\[ qE = qvB \]

We can cancel \( q \) from both sides (since \( q \neq 0 \)):

\[ E = vB \]

### Step 3: Solve for Velocity (v)
Rearranging the equation gives:

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

Substituting the values for \( E \) and \( B \):

- \( B = 0.002 \, \text{Wb/m}^2 \)

\[ v = \frac{3.4 \times 10^5}{0.002} = 1.7 \times 10^8 \, \text{m/s} \]

### Step 4: Calculate the Radius of the Orbit (r)
When the electric field is made zero, the radius of the circular motion of the electrons in the magnetic field can be calculated using the formula:

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

Where:
- \( m = 9.1 \times 10^{-31} \, \text{kg} \) (mass of the electron)
- \( q = 1.6 \times 10^{-19} \, \text{C} \) (charge of the electron)

Substituting the values:

\[ r = \frac{(9.1 \times 10^{-31}) \times (1.7 \times 10^8)}{(1.6 \times 10^{-19}) \times (0.002)} \]

Calculating the numerator:

\[ 9.1 \times 10^{-31} \times 1.7 \times 10^8 = 1.547 \times 10^{-22} \]

Calculating the denominator:

\[ 1.6 \times 10^{-19} \times 0.002 = 3.2 \times 10^{-22} \]

Now substituting back into the radius equation:

\[ r = \frac{1.547 \times 10^{-22}}{3.2 \times 10^{-22}} \approx 0.4834 \, \text{m} \]

### Final Answers
- Velocity of electrons: \( v = 1.7 \times 10^8 \, \text{m/s} \)
- Radius of the orbit: \( r \approx 0.4834 \, \text{m} \)

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