Question

The force per meter between the two wires on a jumper cable being used to start a stalled car is 0.225 N/m. What is the current (A) in the wires, given that they are separated by 2.00 cm

Answer 1

150 A

Step-by-step explanation:

The magnitude of the force per unit length between two current-carrying wires is given by

`(F)/(L)=(\mu_0 I_1 I_2)/(2\pi r)`

where

`\mu_0=4\pi \cdot 10^(-7) H/m` is the vacuum permeability

`I_1,I_2` are the currents in the two wires

r is the distance between the wires

And the force is:

• Attractive if the two wires carry currents in the same direction
• Repulsive if the two wires carry currents in the opposite directions

In this problem, we have:

`(F)/(L)=0.225 N/m` is the force per unit length

`r=2.00 cm = 0.02 m` is the distance between the wires

The two wires carry the same current, so

`I_1=I_2=I`

Therefore we can re-arrange the equation to find the current:

`(F)/(L)=(\mu_0 I^2)/(2\pi r)\\I=\sqrt{(2\pi r (F/L))/(\mu_0)}=\sqrt{(2\pi (0.02)(0.225))/(4\pi \cdot 10^(-7))}=150 A`