Force per unit length acting at one end of each of the two long parallel wires, carrying current I each, kept distance r apart is

Options

A. (a)`(mu_(0))/(4pi)(i^(2))/(r)`
B. (b)`(mu_(0))/(4pi)(2i^(2))/(r)`
C. (c)`(mu_(0))/(4pi)((2i)^(2))/(r)`
D. (d)`(mu_(0))/(4pi)(i^(2))/(4r)`

Answer

To find the force per unit length acting at one end of each of the two long parallel wires carrying current \( I \) and separated by a distance \( r \), we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Setup**:
- We have two long parallel wires, each carrying the same current \( I \).
- The distance between the two wires is \( r \).

2. **Direction of Forces**:
- If the currents in both wires are in the same direction, they will attract each other.
- If the currents are in opposite directions, they will repel each other.

3. **Magnetic Field Due to a Current-Carrying Wire**:
- The magnetic field \( B \) at a distance \( r \) from a long straight wire carrying current \( I \) is given by:
\[ B = \frac{\mu_0 I}{2\pi r} \]
where \( \mu_0 \) is the permeability of free space.

4. **Force on a Current-Carrying Wire in a Magnetic Field**:
- The force \( F \) experienced by a wire of length \( L \) carrying current \( I \) in a magnetic field \( B \) is given by:
\[ F = I \cdot L \cdot B \]

5. **Calculating the Force per Unit Length**:
- For wire 1, the magnetic field at its location due to wire 2 is:
\[ B = \frac{\mu_0 I}{2\pi r} \]
- The force on wire 1 due to the magnetic field created by wire 2 is:
\[ F_1 = I \cdot L \cdot B = I \cdot L \cdot \frac{\mu_0 I}{2\pi r} \]
- The force per unit length \( \frac{F}{L} \) is:
\[ \frac{F_1}{L} = \frac{\mu_0 I^2}{2\pi r} \]

6. **Final Expression**:
- Since both wires experience the same force per unit length, the final expression for the force per unit length acting at one end of each wire is:
\[ \frac{F}{L} = \frac{\mu_0 I^2}{2\pi r} \]

### Final Answer:
The force per unit length acting at one end of each of the two long parallel wires is given by:
\[ \frac{F}{L} = \frac{\mu_0 I^2}{2\pi r} \]