An electron of energy 1800 eV describes a circular path in magnetic field of flux density 0.4 T. The radius of path is `(q = 1.6 xx 10^(-19) C, m_(e)=9.1 xx 10^(-31) kg)`

Options

A. `2.58xx10^(-4)m`
B. `3.58xx10^(-4)m`
C. `2.58xx10^(-3)m`
D. `3.58xx10^(-4)m`

Explanation

To solve the problem of finding the radius of the circular path of an electron moving in a magnetic field, we can follow these steps:

### Step 1: Understand the Forces
The centripetal force required to keep the electron in circular motion is provided by the magnetic force. Therefore, we can equate the centripetal force \( F_c \) to the magnetic force \( F_m \).

\[ F_c = F_m \]

### Step 2: Write the Expressions for Forces
The centripetal force can be expressed as:

\[ F_c = \frac{mv^2}{R} \]

where:
- \( m \) is the mass of the electron,
- \( v \) is the velocity of the electron,
- \( R \) is the radius of the circular path.

The magnetic force on a charged particle moving in a magnetic field is given by:

\[ F_m = qvB \]

where:
- \( q \) is the charge of the electron,
- \( B \) is the magnetic flux density.

### Step 3: Set the Forces Equal
Setting the two expressions equal gives us:

\[ \frac{mv^2}{R} = qvB \]

### Step 4: Solve for Radius \( R \)
We can simplify this equation to solve for \( R \):

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

### Step 5: Relate Kinetic Energy to Velocity
The kinetic energy \( KE \) of the electron is given by:

\[ KE = \frac{1}{2}mv^2 \]

Given that the energy of the electron is 1800 eV, we can convert this energy into joules:

\[ KE = 1800 \, \text{eV} \times 1.6 \times 10^{-19} \, \text{J/eV} = 2.88 \times 10^{-16} \, \text{J} \]

From the kinetic energy expression, we can solve for \( v \):

\[ v = \sqrt{\frac{2KE}{m}} = \sqrt{\frac{2 \times 2.88 \times 10^{-16}}{9.1 \times 10^{-31}}} \]

### Step 6: Substitute \( v \) Back into the Radius Equation
Now we can substitute \( v \) back into the equation for \( R \):

\[ R = \frac{m}{qB} \cdot \sqrt{\frac{2KE}{m}} = \frac{1}{qB} \cdot \sqrt{2m \cdot KE} \]

### Step 7: Plug in the Values
Now we can substitute the known values:

- \( m = 9.1 \times 10^{-31} \, \text{kg} \)
- \( q = 1.6 \times 10^{-19} \, \text{C} \)
- \( B = 0.4 \, \text{T} \)
- \( KE = 2.88 \times 10^{-16} \, \text{J} \)

Substituting these values gives:

\[ R = \frac{1}{(1.6 \times 10^{-19})(0.4)} \cdot \sqrt{2(9.1 \times 10^{-31})(2.88 \times 10^{-16})} \]

### Step 8: Calculate the Radius
After performing the calculations, we find:

\[ R \approx 3.58 \times 10^{-4} \, \text{m} \]

Thus, the radius of the path is approximately \( 3.58 \times 10^{-4} \, \text{m} \).

### Final Answer
The radius of the circular path is \( R \approx 3.58 \times 10^{-4} \, \text{m} \).

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