Question

Two infinite line charges parallel to each other are moving with a constant velocity $v$ in the same direction as shown in the figure. The separation between two line charges is $d$. The magnetic attraction balances the electric repulsion when, [e = speed of light in free space]

• A. v= √2 e
• B. v= e/√2
• C. v=e
• D. v-e/2

Answer 1

The Correct Option is C

Given: Two infinite line charges moving with velocity \( v \) in the same direction, separated by a distance \( d \). We are to find the condition when magnetic attraction balances electric repulsion. Let \( e \) be the speed of light in free space.

Concepts involved: - A moving line charge produces both electric and magnetic fields. 
- Electric field causes repulsion between like charges. 
- Magnetic field due to current causes attraction (since currents are in same direction). 
- At equilibrium: Electric force = Magnetic force Electric Force per unit length: If linear charge density is \( \lambda \), then: \[ F_E = \frac{1}{2\pi \varepsilon_0} \cdot \frac{\lambda^2}{d} \]

 Magnetic Force per unit length: Moving line charges act like current-carrying wires with current \( I = \lambda v \), so: \[ F_B = \frac{\mu_0}{2\pi} \cdot \frac{I^2}{d} = \frac{\mu_0}{2\pi} \cdot \frac{\lambda^2 v^2}{d} \] 

Equating: \[ F_E = F_B \Rightarrow \frac{1}{2\pi \varepsilon_0} \cdot \frac{\lambda^2}{d} = \frac{\mu_0}{2\pi} \cdot \frac{\lambda^2 v^2}{d} \] Cancel \( \lambda^2 \), \( d \), \( 2\pi \): \[ \frac{1}{\varepsilon_0} = \mu_0 v^2 \Rightarrow v^2 = \frac{1}{\mu_0 \varepsilon_0} \Rightarrow v = \frac{1}{\sqrt{\mu_0 \varepsilon_0}} = e \] 

Final Answer: \( v = e \)