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

Concept: When two long, straight, parallel current-carrying conductors are placed close to each other, they exert a magnetic force on one another. The magnitude of the force between two parallel conductors is given by \[ F=\frac{\mu_{0}I_{1}I_{2}l}{2\pi d}, \] where \[ \mu_{0}=4\pi\times10^{-7}\;H/m \] is the permeability of free space, \[ I_{1},\,I_{2}=\text{Currents in the two conductors}, \] \[ l=\text{Length of each conductor}, \] \[ d=\text{Perpendicular distance between the conductors}. \] The direction of the force depends upon the direction of current:

• Currents in the

same direction $\Rightarrow$ Attraction.

• Currents in the

opposite directions $\Rightarrow$ Repulsion.
Since the currents are flowing in opposite directions, the wires repel each other.

Step 1: Write the given data.
Given, \[ I_{1}=I_{2}=2\,A, \] \[ l=0.5\,m, \] \[ d=10\,cm=0.1\,m. \] Also, \[ \mu_{0}=4\pi\times10^{-7}\;H/m. \]

Step 2: Write the formula for the force between two parallel conductors.
The force is \[ F=\frac{\mu_{0}I_{1}I_{2}l}{2\pi d}. \] Substituting the given values, \[ F= \frac{(4\pi\times10^{-7})(2)(2)(0.5)} {2\pi(0.1)}. \]

Step 3: Simplify the numerator.
Since, \[ (2)(2)=4, \] and \[ 4\times0.5=2, \] therefore, \[ F= \frac{(4\pi\times10^{-7})\times2} {2\pi\times0.1}. \] Hence, \[ F= \frac{8\pi\times10^{-7}} {0.2\pi}. \] The factor \(\pi\) cancels from numerator and denominator.

Step 4: Calculate the force.
Thus, \[ F= \frac{8\times10^{-7}} {0.2} = 40\times10^{-7} = 4\times10^{-6}\;N. \] Therefore, \[ \boxed{F=4\times10^{-6}\;N.} \] Since the currents are in opposite directions, the force is \[ \boxed{\text{Repulsive}.} \] Hence, the correct answer is \[ \boxed{\textbf{Option (B)}}. \]