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
Show Hint
Force \(\propto \frac{I^2}{d}\), so check both current and distance changes.
Step 1: Understanding the Concept:
Two parallel current-carrying wires exert a magnetic force on each other. The force per unit length depends on the product of the currents and is inversely proportional to the separation distance. Step 2: Key Formula or Approach:
The force \(F\) on a length \(L\) of a wire is:
\[ F = \frac{\mu_0 I_1 I_2 L}{2\pi d} \]
Since initial currents are equal: \(F \propto \frac{I^2}{d}\). Step 3: Detailed Explanation:
Let the initial force be \(F \propto \frac{I^2}{d}\).
According to the problem, the new conditions are:
1. Currents are doubled: \(I'_1 = 2I\), \(I'_2 = 2I\).
2. Distance is halved: \(d' = d/2\).
Let the new force be \(F_2\):
\[ F_2 = \frac{\mu_0 (2I) (2I) L}{2\pi (d/2)} \]
Factor out the constants:
\[ F_2 = \frac{\mu_0 \cdot 4I^2 \cdot L}{2\pi d / 2} = \frac{\mu_0 \cdot 4I^2 \cdot L \cdot 2}{2\pi d} \]
\[ F_2 = 8 \times \left( \frac{\mu_0 I^2 L}{2\pi d} \right) \]
\[ F_2 = 8F \]
Step 4: Final Answer:
The new force \(F_2\) will be \(8F\).