A long straight wire is kept along x-axis. It carries a current i in the positive x-direction. A proton and an electron are placed at (0, a,0) and (0, -a, 0) respecitively. The proton is imparted an initial velocity v along +z axis and the electron is imparted an initial velocity v along +x-axis. The magnetic forces experienced by the two particles at the instant are

Options

A. `(mu_(0)iev)/(2pia) hat(i) (mu_(0)iev)/(2pia) hat(i)`
B. `0, (-mu_(0)iev)/(2pia) hat(j)`
C. `0, (mu_(0)iev)/(2pia) hat(j)`
D. `0, 0`

Answer

To solve the problem, we need to determine the magnetic forces experienced by a proton and an electron placed in a magnetic field created by a long straight wire carrying a current.

### Step-by-Step Solution:

1. **Identify the Magnetic Field due to the Wire**:
The magnetic field \( \mathbf{B} \) at a distance \( r \) from a long straight wire carrying a current \( I \) is given by the formula:
\[ \mathbf{B} = \frac{\mu_0 I}{2 \pi r} \hat{\phi} \]
where \( \hat{\phi} \) is the azimuthal unit vector that follows the right-hand rule. For our wire along the x-axis, at point \( (0, a, 0) \) (where the proton is located), the distance \( r = a \) and the magnetic field direction will be in the negative z-direction (since the current is in the positive x-direction).

Thus, the magnetic field at the position of the proton is:
\[ \mathbf{B} = -\frac{\mu_0 I}{2 \pi a} \hat{k} \]

2. **Calculate the Magnetic Force on the Proton**:
The magnetic force \( \mathbf{F} \) on a charged particle moving in a magnetic field is given by:
\[ \mathbf{F} = q(\mathbf{v} \times \mathbf{B}) \]
For the proton, with charge \( q_p = +e \) and velocity \( \mathbf{v}_p = v \hat{k} \):
\[ \mathbf{F}_p = e(v \hat{k} \times -\frac{\mu_0 I}{2 \pi a} \hat{k}) \]
The cross product \( \hat{k} \times \hat{k} = 0 \), hence:
\[ \mathbf{F}_p = 0 \]

3. **Calculate the Magnetic Force on the Electron**:
For the electron, with charge \( q_e = -e \) and velocity \( \mathbf{v}_e = v \hat{i} \):
\[ \mathbf{F}_e = -e(v \hat{i} \times -\frac{\mu_0 I}{2 \pi a} \hat{k}) \]
Now, we calculate the cross product:
\[ \hat{i} \times \hat{k} = \hat{j} \]
Therefore:
\[ \mathbf{F}_e = -e \left(v \left(-\frac{\mu_0 I}{2 \pi a}\right) \hat{j}\right) = \frac{\mu_0 I e v}{2 \pi a} \hat{j} \]

4. **Final Results**:
- The magnetic force on the proton is \( \mathbf{F}_p = 0 \).
- The magnetic force on the electron is \( \mathbf{F}_e = \frac{\mu_0 I e v}{2 \pi a} \hat{j} \).