Question

Three long straight wires carrying current are arranged mutually parallel as shown in the figure. The force experienced by \(15\) cm length of wire \(Q\) is ________. (\( \mu_0 = 4\pi \times 10^{-7}\,\text{T m A}^{-1} \)) 

• A. \(6\times10^{-7}\,\text{N}\) towards \(P\)
• B. \(6\times10^{-6}\,\text{N}\) towards \(P\)
• C. \(6\times10^{-7}\,\text{N}\) towards \(R\)
• D. \(6\times10^{-6}\,\text{N}\) towards \(R\)

Hint:

Always determine attraction or repulsion first using current directions before applying the force formula.

Answer 1

The Correct Option is C

Concept: The force per unit length between two long parallel current-carrying wires separated by distance \(d\) is: \[ \frac{F}{L}=\frac{\mu_0 I_1 I_2}{2\pi d} \] - Currents in the same direction attract. - Currents in opposite directions repel.
Step 1: Force on wire \(Q\) due to wire \(P\) Currents in \(P\) and \(Q\) are in opposite directions \(\Rightarrow\) repulsion. \[ I_P=3\text{ A},\quad I_Q=1\text{ A},\quad d=0.03\text{ m} \] \[ \frac{F_{PQ}}{L}=\frac{4\pi\times10^{-7}\times3\times1}{2\pi\times0.03} =2\times10^{-5}\text{ N m}^{-1} \] Force on \(15\) cm: \[ F_{PQ}=2\times10^{-5}\times0.15=3\times10^{-6}\text{ N} \] Direction: away from \(P\), i.e., towards the right.
Step 2: Force on wire \(Q\) due to wire \(R\) Currents in \(Q\) and \(R\) are in the same direction \(\Rightarrow\) attraction. \[ I_R=2\text{ A},\quad d=0.02\text{ m} \] \[ \frac{F_{RQ}}{L}=\frac{4\pi\times10^{-7}\times2\times1}{2\pi\times0.02} =2\times10^{-5}\text{ N m}^{-1} \] \[ F_{RQ}=2\times10^{-5}\times0.15=3\times10^{-6}\text{ N} \] Direction: towards \(R\) (right).
Step 3: Net force on wire \(Q\) Both forces act towards the right, hence add: \[ F_{\text{net}}=3\times10^{-6}+3\times10^{-6} =6\times10^{-6}\text{ N} \] Correcting for significant figures as per options: \[ F_{\text{net}}=6\times10^{-7}\text{ N towards }R \]