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 analyze the situation step by step, focusing on the magnetic forces acting on the proton and the electron due to the current-carrying wire.

### Step 1: Identify the Magnetic Field
The long straight wire carrying current \( I \) in the positive x-direction generates a magnetic field around it. The magnetic field \( \vec{B} \) at a distance \( a \) from the wire (along the y-axis) can be calculated using Ampère's law:

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

The direction of the magnetic field can be determined using the right-hand rule. For a wire carrying current in the positive x-direction, the magnetic field at the point (0, a, 0) (where the proton is located) will be directed out of the page (positive z-direction), and at the point (0, -a, 0) (where the electron is located), it will be directed into the page (negative z-direction).

### Step 2: Analyze the Proton's Motion
The proton is located at (0, a, 0) and is given an initial velocity \( \vec{v} \) along the positive z-axis.

- **Velocity of Proton**: \( \vec{v}_{p} = v \hat{k} \)
- **Magnetic Field at Proton's Location**: \( \vec{B} = B \hat{k} \) (where \( B = \frac{\mu_0 I}{2 \pi a} \))

The magnetic force \( \vec{F} \) on a charged particle is given by the equation:

\[ \vec{F} = q \vec{v} \times \vec{B} \]

For the proton, substituting the values:

\[ \vec{F}_{p} = q_{p} \vec{v}_{p} \times \vec{B} = e \cdot (v \hat{k}) \times \left(\frac{\mu_0 I}{2 \pi a} \hat{k}\right) \]

Since \( \vec{v}_{p} \) and \( \vec{B} \) are in the same direction (both along \( \hat{k} \)), the cross product \( \vec{v}_{p} \times \vec{B} = 0 \).

Thus, the magnetic force on the proton is:

\[ \vec{F}_{p} = 0 \]

### Step 3: Analyze the Electron's Motion
The electron is located at (0, -a, 0) and is given an initial velocity \( \vec{v} \) along the positive x-axis.

- **Velocity of Electron**: \( \vec{v}_{e} = v \hat{i} \)
- **Magnetic Field at Electron's Location**: \( \vec{B} = -B \hat{k} \) (into the page)

Now, we calculate the magnetic force on the electron:

\[ \vec{F}_{e} = q_{e} \vec{v}_{e} \times \vec{B} = -e \cdot (v \hat{i}) \times \left(-\frac{\mu_0 I}{2 \pi a} \hat{k}\right) \]

Calculating the cross product:

\[ \vec{F}_{e} = -e \cdot v \cdot \left(-\frac{\mu_0 I}{2 \pi a}\right) (\hat{i} \times -\hat{k}) = e \cdot v \cdot \frac{\mu_0 I}{2 \pi a} (\hat{j}) \]

The direction of \( \hat{i} \times -\hat{k} \) is \( \hat{j} \) (positive y-direction). Therefore, the force on the electron is:

\[ \vec{F}_{e} = \frac{e v \mu_0 I}{2 \pi a} \hat{j} \]

### Conclusion
- The magnetic force on the proton is \( \vec{F}_{p} = 0 \).
- The magnetic force on the electron is \( \vec{F}_{e} = \frac{e v \mu_0 I}{2 \pi a} \hat{j} \).

### Final Answer
The magnetic forces experienced by the two particles at the instant are:
- Proton: \( 0 \)
- Electron: \( \frac{e v \mu_0 I}{2 \pi a} \hat{j} \)

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