Question

Two long straight wires A and B carrying equal current 'I' were kept parallel to each other at distance ' d ' apart. Magnitude of magnetic force experienced by length \(L\) of wire A is ' \(F\) '. If the distance between the wires is made half and currents are doubled, force \(F_2\) on length \(L\) of wire A will be

• A. \(2F\)
• B. \(F\)
• C. \(8F\)
• D. \(4F\)

Hint:

Force \(\propto \frac{I^2}{d}\), so check both current and distance changes.

Answer 1

The Correct Option is C

Concept:
Force between parallel wires: \[ F \propto \frac{I_1 I_2}{d} \] Step 1: Initial condition. \[ F \propto \frac{I^2}{d} \]
Step 2: New condition. \[ I \rightarrow 2I, \quad d \rightarrow \frac{d}{2} \] \[ F_2 \propto \frac{(2I)(2I)}{d/2} = \frac{4I^2}{d/2} \] \[ F_2 \propto \frac{4I^2 \times 2}{d} = \frac{8I^2}{d} \]
Step 3: Compare. \[ F_2 = 8F \]
Step 4: Conclusion. Force becomes \(8F\)