A straight conductor carries a current of `5A`. An electron travelling with a speed of `5xx10^(6)ms^(-1)` parallel to the wire at a distance of 0.1m from the conductor, experiences a force of

Options

A. (a)`8xx10^(-20)N`
B. (b)`3.2xx10^(-19)N`
C. (c)`8xx10^(-18)N`
D. (d)`1.6xx10^(-19)N`

Answer

To solve the problem, we need to calculate the force experienced by an electron moving parallel to a straight conductor carrying a current. Here are the steps to find the solution:

### Step 1: Identify the given values
- Current (I) = 5 A
- Velocity of the electron (v) = \(5 \times 10^6 \, \text{m/s}\)
- Distance from the conductor (d) = 0.1 m
- Charge of the electron (q) = \(1.6 \times 10^{-19} \, \text{C}\)

### Step 2: Calculate the magnetic field (B) created by the straight conductor
The magnetic field (B) at a distance (d) from a long straight conductor carrying current (I) is given by the formula:
\[ B = \frac{\mu_0 I}{2 \pi d} \]
Where:
- \(\mu_0\) (permeability of free space) = \(4\pi \times 10^{-7} \, \text{T m/A}\)

Substituting the values:
\[ B = \frac{(4\pi \times 10^{-7}) \times 5}{2\pi \times 0.1} \]
Simplifying:
\[ B = \frac{(4 \times 5) \times 10^{-7}}{2 \times 0.1} = \frac{20 \times 10^{-7}}{0.2} = 10^{-6} \, \text{T} \]

### Step 3: Calculate the force (F) on the electron
The force experienced by a charged particle moving in a magnetic field is given by:
\[ F = q v B \]
Substituting the values:
\[ F = (1.6 \times 10^{-19}) \times (5 \times 10^6) \times (10^{-6}) \]
Calculating:
\[ F = 1.6 \times 5 \times 10^{-19} \times 10^0 = 8 \times 10^{-19} \, \text{N} \]

### Final Answer
The force experienced by the electron is:
\[ F = 8 \times 10^{-19} \, \text{N} \]