Two parallel beams of protons and electrons, carrying equal currents are fixed at a separation d. The protons and electrons move in opposite directions. P is a point on a line joining the beams , at a distacne x from any one beam. The magnetic field at P is B . If B is plotted against x, which of the following best represents the resulting curve

Options

A.
B. `##DPP_PHY_MEI_2_1_E01_099_O02.png" width="30%">
C.
D.

Answer

To solve the problem, we need to analyze the magnetic field produced by two parallel beams of protons and electrons that are carrying equal currents and moving in opposite directions. Let's break down the solution step by step.

### Step 1: Understand the Configuration
We have two parallel beams:
- Beam 1: Protons moving upwards (current in the same direction).
- Beam 2: Electrons moving downwards (current in the opposite direction).

The distance between the two beams is \( D \).

### Step 2: Identify the Point of Interest
Let \( P \) be a point on the line joining the two beams, at a distance \( x \) from the proton beam. Therefore, the distance from the electron beam to point \( P \) is \( D - x \).

### Step 3: Magnetic Field Due to Each Beam
The magnetic field \( B \) due to a long straight current-carrying wire at a distance \( r \) is given by the formula:
\[ B = \frac{\mu_0 I}{2 \pi r} \]
where \( \mu_0 \) is the permeability of free space, and \( I \) is the current.

- **Magnetic Field due to Protons at Point P**:
\[ B_P = \frac{\mu_0 I}{2 \pi x} \]
- **Magnetic Field due to Electrons at Point P**:
\[ B_E = \frac{\mu_0 I}{2 \pi (D - x)} \]

### Step 4: Determine the Net Magnetic Field at Point P
Since the protons and electrons are moving in opposite directions, the magnetic fields will have opposite signs. Therefore, the net magnetic field \( B \) at point \( P \) is:
\[ B = B_P - B_E = \frac{\mu_0 I}{2 \pi x} - \frac{\mu_0 I}{2 \pi (D - x)} \]
Factoring out the common terms gives:
\[ B = \frac{\mu_0 I}{2 \pi} \left( \frac{1}{x} - \frac{1}{D - x} \right) \]

### Step 5: Analyze the Behavior of B as x Changes
1. **At \( x = D/2 \)**:
- The magnetic field \( B \) becomes zero because the contributions from both beams cancel each other out.

2. **For \( x < D/2 \)**:
- The term \( \frac{1}{x} \) dominates, leading to \( B > 0 \).

3. **For \( x > D/2 \)**:
- The term \( \frac{1}{D - x} \) dominates, leading to \( B < 0 \).

### Step 6: Plotting B against x
- As \( x \) approaches \( 0 \), \( B \) approaches \( +\infty \).
- At \( x = D/2 \), \( B = 0 \).
- As \( x \) approaches \( D \), \( B \) approaches \( -\infty \).

This behavior indicates that the graph of \( B \) versus \( x \) will show a positive value for \( x < D/2 \), zero at \( x = D/2 \), and negative for \( x > D/2 \).

### Conclusion
The resulting curve of \( B \) plotted against \( x \) will look like a hyperbola that approaches infinity as \( x \) approaches \( 0 \) and approaches negative infinity as \( x \) approaches \( D \), crossing the x-axis at \( D/2 \).