Electric charge `q` is distributed uniformly over a rod of length `l`. The rod is placed parallel to a long wire carrying a current `i`. The separation between the rod and the wire is a. The force needed to move the rod along its length with a uniform velocity `v` is

Options

A. `(mu_0iqv)/(2pia)`
B. `(mu_0iqv)/(4pia)`
C. `(mu_0iqvl)/(2pia)`
D. `(mu_0iqvl)/(4pia)`

Explanation

To solve the problem, we need to determine the force required to move a charged rod parallel to a long wire carrying a current. Here’s a step-by-step solution:

### Step 1: Understand the Setup
- We have a rod of length \( l \) with a uniform charge \( q \) distributed along its length.
- The rod is placed parallel to a long straight wire carrying a current \( i \).
- The separation between the rod and the wire is \( a \).

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

### Step 3: Calculate the Force on the Charged Rod
- The force \( F \) on a charge \( q \) moving with a velocity \( v \) in a magnetic field \( B \) is given by:
\[ F = q v B \sin \theta \]
where \( \theta \) is the angle between the velocity vector and the magnetic field vector.
- In this case, since the rod is moving parallel to the magnetic field, \( \theta = 90^\circ \) and \( \sin 90^\circ = 1\). Therefore, we can simplify the equation to:
\[ F = q v B \]

### Step 4: Substitute the Expression for Magnetic Field
- Now we substitute the expression for \( B \) from Step 2 into the force equation:
\[ F = q v \left( \frac{\mu_0 I}{2 \pi a} \right) \]

### Step 5: Final Expression for the Force
- The force required to move the rod with a uniform velocity \( v \) is:
\[ F = \frac{\mu_0 q I v}{2 \pi a} \]

### Conclusion
- The force needed to move the rod along its length with a uniform velocity \( v \) is given by:
\[ F = \frac{\mu_0 q I v}{2 \pi a} \]