Question

Two parallel long wires carry the same current and repel each other with a force F per unit length. If both these currents are doubled and the wire separation tripled, the force per unit length becomes:

Answer 1

Answer:

When the currents are doubled and the wire separation tripled, the force per unit length becomes 4/3 of the initial force per unit length.

Explanation:

Since the two wires carry the same current, the force F of repulsion per unit length is given by:

$\frac{F}{l} = \frac{I^{2}B\mu_{0}}{2\pi h}$    

Where:

I: is the current

B: is the magnetic field

l: is the wire's length

h: is the separation between the wires

μ₀: is the permeability constant  

When the two parallel long wires carry the same current and repel each other with a force F per unit length we have:

$\frac{F_{1}}{l} = \frac{I^{2}B\mu_{0}}{2\pi h}$

And when the currents are doubled and the wire separation tripled, the force per unit length becomes:

$\frac{F_{2}}{l} = \frac{(2I)^{2}B\mu_{0}}{2\pi (3h)}$

$\frac{F_{2}}{l} = \frac{4I^{2}B\mu_{0}}{2\pi (3h)}$  

$\frac{F_{2}}{l} = \frac{4}{3}\frac{I^{2}B\mu_{0}}{2\pi h} = \frac{4}{3}\frac{F_{1}}{l}$  

Therefore, when the currents are doubled and the wire separation tripled, the force per unit length becomes 4/3 of the initial force per unit length.

I hope it helps you!