A long straight wire carries a current of `2A`. An electron travels with a velocity of `4xx10^4ms^-1` parallel to the wire `0*1mm` from it, and in a direction opposite to the current. What force does the magnetic field of current exert on the moving electron. Charge on electron `=1*6xx10^(-19)C`.

Explanation

To solve the problem, we will follow these steps:

### Step 1: Calculate the Magnetic Field (B) due to the Current-Carrying Wire
The formula for the magnetic field around a long straight wire carrying current \( I \) at a distance \( d \) is given by:

\[ B = \frac{\mu_0 I}{2 \pi d} \]

Where:
- \( \mu_0 = 4\pi \times 10^{-7} \, \text{T m/A} \) (permeability of free space)
- \( I = 2 \, \text{A} \) (current in the wire)
- \( d = 0.1 \, \text{mm} = 0.1 \times 10^{-3} \, \text{m} = 1 \times 10^{-4} \, \text{m} \)

Substituting the values:

\[ B = \frac{4\pi \times 10^{-7} \times 2}{2 \pi \times 1 \times 10^{-4}} = \frac{8 \times 10^{-7}}{2 \times 10^{-4}} = 4 \times 10^{-3} \, \text{T} \]

### Step 2: Determine the Direction of the Magnetic Field
Using the right-hand rule, if the current is flowing in one direction, the magnetic field circles around the wire. Since the electron is moving opposite to the current, the magnetic field at the position of the electron will be directed out of the plane (positive \( k \)-direction).

### Step 3: Calculate the Force (F) on the Electron
The force on a charged particle moving in a magnetic field is given by:

\[ F = q(v \times B) \]

Where:
- \( q = -1.6 \times 10^{-19} \, \text{C} \) (charge of the electron)
- \( v = 4 \times 10^4 \, \text{m/s} \) (velocity of the electron)
- \( B = 4 \times 10^{-3} \, \text{T} \) (magnetic field calculated above)

### Step 4: Determine the Velocity and Magnetic Field Vectors
Assuming the electron is moving in the negative x-direction:

\[ \vec{v} = -4 \times 10^4 \hat{i} \, \text{m/s} \]
\[ \vec{B} = 4 \times 10^{-3} \hat{k} \, \text{T} \]

### Step 5: Calculate the Cross Product \( v \times B \)
Using the determinant method for the cross product:

\[ \vec{F} = q \vec{v} \times \vec{B} = -1.6 \times 10^{-19} \hat{q} \cdot \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -4 \times 10^4 & 0 & 0 \\ 0 & 0 & 4 \times 10^{-3} \end{vmatrix} \]

Calculating the determinant:

\[ \vec{v} \times \vec{B} = \hat{i}(0 \cdot 4 \times 10^{-3} - 0) - \hat{j}(-4 \times 10^4 \cdot 4 \times 10^{-3} - 0) + \hat{k}(0 - 0) \]
\[ = 0 \hat{i} + 0 \hat{j} + (-4 \times 10^4 \cdot 4 \times 10^{-3}) \hat{j} \]
\[ = 16 \hat{j} \, \text{(in the positive y-direction)} \]

### Step 6: Calculate the Force Magnitude and Direction
Now substituting back into the force equation:

\[ F = -1.6 \times 10^{-19} \cdot 16 = -2.56 \times 10^{-18} \, \text{N} \]

The negative sign indicates that the force is in the negative y-direction.

### Final Answer
The force exerted on the moving electron by the magnetic field of the current is:

\[ F = 2.56 \times 10^{-18} \, \text{N} \, \text{(in the negative y-direction)} \]

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