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 find the radius of the circular path described by an electron accelerated through a potential difference of 500 V and entering a magnetic field of 100 mT, we can follow these steps:

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

Substituting the values:
\[ KE = (1.6 \times 10^{-19} \, \text{C})(500 \, \text{V}) = 8.0 \times 10^{-17} \, \text{J} \]

### Step 2: Relate Kinetic Energy to Velocity
The kinetic energy can also be expressed in terms of the mass \( m \) and velocity \( v \) of the electron:
\[ KE = \frac{1}{2} mv^2 \]
Setting the two expressions for kinetic energy equal gives:
\[ \frac{1}{2} mv^2 = eV \]
From this, we can solve for \( v \):
\[ v = \sqrt{\frac{2eV}{m}} \]

### Step 3: Substitute the Mass of the Electron
The mass of the electron \( m \) is approximately \( 9.1 \times 10^{-31} \, \text{kg} \). Now substituting the known values into the velocity equation:
\[ v = \sqrt{\frac{2(1.6 \times 10^{-19} \, \text{C})(500 \, \text{V})}{9.1 \times 10^{-31} \, \text{kg}}} \]
Calculating the value:
\[ v = \sqrt{\frac{1.6 \times 10^{-19} \times 1000}{9.1 \times 10^{-31}}} = \sqrt{\frac{1.6 \times 10^{-16}}{9.1 \times 10^{-31}}} \]
\[ v \approx \sqrt{1.75824176 \times 10^{14}} \approx 1.32 \times 10^7 \, \text{m/s} \]

### Step 4: 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:
- \( B = 100 \, \text{mT} = 100 \times 10^{-3} \, \text{T} = 0.1 \, \text{T} \)

Substituting the values:
\[ r = \frac{(9.1 \times 10^{-31} \, \text{kg})(1.32 \times 10^7 \, \text{m/s})}{(1.6 \times 10^{-19} \, \text{C})(0.1 \, \text{T})} \]
Calculating the numerator:
\[ = 1.1992 \times 10^{-23} \, \text{kg m/s} \]
Calculating the denominator:
\[ = 1.6 \times 10^{-20} \, \text{C T} \]
Now calculating \( r \):
\[ r = \frac{1.1992 \times 10^{-23}}{1.6 \times 10^{-20}} \approx 7.49 \times 10^{-4} \, \text{m} \approx 0.749 \, \text{mm} \]

### Final Answer
The radius of the circular path described by the electron is nearly:
\[ r \approx 7.49 \times 10^{-4} \, \text{m} \quad \text{or} \quad 0.749 \, \text{mm} \]