An electron accelerated through 500 V , enters a transverse uniform magnetic field of magnitude 100 mT . The radius of the circular path described by the electron is nearly

Options

A. `7.54xx10^(-1)m`
B. `7.54xx10^(-2)m`
C. `7.54xx10^(-3)m`
D. `7.54xx10^(-4)m`

Answer

To solve the problem, we need to find the radius of the circular path described by an electron that has been accelerated through a potential difference of 500 V and then enters a magnetic field of 100 mT. Here’s a step-by-step solution:

### Step 1: Calculate the velocity of the electron after acceleration
The kinetic energy (KE) gained by the electron when it is accelerated through a potential difference (V) is given by the equation:
\[ KE = eV \]
where:
- \( e \) is the charge of the electron (\( 1.6 \times 10^{-19} \) C)
- \( V \) is the potential difference (500 V)

The kinetic energy can also be expressed in terms of mass and velocity:
\[ KE = \frac{1}{2} mv^2 \]
where:
- \( m \) is the mass of the electron (\( 9.1 \times 10^{-31} \) kg)
- \( v \) is the velocity of the electron

Setting these two expressions for kinetic energy equal to each other:
\[ eV = \frac{1}{2} mv^2 \]
Rearranging for \( v \):
\[ v = \sqrt{\frac{2eV}{m}} \]

### Step 2: Substitute the values to find the velocity
Substituting the known values:
\[ v = \sqrt{\frac{2 \times (1.6 \times 10^{-19} \, \text{C}) \times (500 \, \text{V})}{9.1 \times 10^{-31} \, \text{kg}}} \]

Calculating the numerator:
\[ 2 \times (1.6 \times 10^{-19}) \times 500 = 1.6 \times 10^{-16} \, \text{J} \]

Now, substituting this into the equation for \( v \):
\[ v = \sqrt{\frac{1.6 \times 10^{-16}}{9.1 \times 10^{-31}}} \]

Calculating this gives:
\[ v \approx \sqrt{1.758241 \times 10^{14}} \approx 1.33 \times 10^{7} \, \text{m/s} \]

### Step 3: Calculate the radius of the circular path
The radius \( r \) of the circular path in a magnetic field is given by:
\[ r = \frac{mv}{eB} \]
where:
- \( B \) is the magnetic field strength (100 mT = \( 100 \times 10^{-3} \, \text{T} \))

Substituting the known values:
\[ r = \frac{(9.1 \times 10^{-31} \, \text{kg}) \times (1.33 \times 10^{7} \, \text{m/s})}{(1.6 \times 10^{-19} \, \text{C}) \times (100 \times 10^{-3} \, \text{T})} \]

Calculating the denominator:
\[ (1.6 \times 10^{-19}) \times (100 \times 10^{-3}) = 1.6 \times 10^{-21} \, \text{C} \cdot \text{T} \]

Now substituting back into the equation for \( r \):
\[ r = \frac{(9.1 \times 10^{-31}) \times (1.33 \times 10^{7})}{1.6 \times 10^{-21}} \]

Calculating this gives:
\[ r \approx \frac{1.2093 \times 10^{-23}}{1.6 \times 10^{-21}} \approx 7.55 \times 10^{-3} \, \text{m} \]

### Final Answer
Thus, the radius of the circular path described by the electron is approximately:
\[ r \approx 7.54 \times 10^{-4} \, \text{m} \]