Question

Two long, straight, parallel conductors carry the same current and attract each other with a force F per unit length. If the current is doubled in each conductor and their separation is made half, the force per unit length becomes:

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

Hint:

The force per unit length is proportional to I1 I2 by d. Doubling both currents gives a factor 4 and halving d gives a factor 2.

Answer 1

The Correct Option is D

Step 1: Understanding the concept:
Two long parallel wires with currents \(I_1\) and \(I_2\) at separation \(d\) exert a force per unit length \(\dfrac{\mu_0 I_1I_2}{2\pi d}\) on each other.

Step 2: Original force:
With equal currents \(I\): \(F = \dfrac{\mu_0 I^{2}}{2\pi d}\).

Step 3: New force:
The current becomes \(2I\) in each wire, and the separation becomes \(d/2\).
\[ F_{n} = \frac{\mu_0 (2I)(2I)}{2\pi (d/2)} = \frac{\mu_0\times 4I^{2}\times 2}{2\pi d} = 8\times\frac{\mu_0 I^{2}}{2\pi d} = 8F \]

Step 4: Check the options:
(A) \(F/2\): wrong, this treats the force as growing with distance.
(B) 2F: wrong, this counts only one of the two changes.
(C) 4F: wrong, this counts the current change \(2\times 2\) but ignores the halved separation.
(D) 8F: correct.

Final Answer:
The new force per unit length is 8F. The answer is option (D). \[ \boxed{F_{n} = 8F} \]