Question

Two parallel wires carry equal currents of \(2\,A\) in opposite directions. If the length of each wire is \(0.5\,m\) and the distance between them is \(10\,cm\), then find the force between them.

• A. \(2\times10^{-6}\,N\)
• B. \(4\times10^{-6}\,N\)
• C. \(8\times10^{-6}\,N\)
• D. \(1.6\times10^{-5}\,N\)

Hint:

Always remember the force between two parallel current-carrying conductors: \[ \boxed{F=\frac{\mu_{0}I_{1}I_{2}l}{2\pi d}} \] where \(d\) is the separation between the wires. Also remember: \[ \boxed{ \begin{aligned} \text{Same direction currents} &\Rightarrow \text{Attraction},\\ \text{Opposite direction currents} &\Rightarrow \text{Repulsion}. \end{aligned} } \] Before substituting values, always convert the distance into SI units (metres).

Answer 1

The Correct Option is B

The force per unit length between two parallel current-carrying wires is $F/L = \mu_0 I_1 I_2/(2\pi d)$. Since the currents flow in opposite directions, by the right-hand rule the magnetic forces on each wire are repulsive.