Answer:
150 A
Explanation:
The magnitude of the force per unit length between two current-carrying wires is given by
[tex]\frac{F}{L}=\frac{\mu_0 I_1 I_2}{2\pi r}[/tex]
where
[tex]\mu_0=4\pi \cdot 10^{-7} H/m[/tex] is the vacuum permeability
[tex]I_1,I_2[/tex] are the currents in the two wires
r is the distance between the wires
And the force is:
In this problem, we have:
[tex]\frac{F}{L}=0.225 N/m[/tex] is the force per unit length
[tex]r=2.00 cm = 0.02 m[/tex] is the distance between the wires
The two wires carry the same current, so
[tex]I_1=I_2=I[/tex]
Therefore we can re-arrange the equation to find the current:
[tex]\frac{F}{L}=\frac{\mu_0 I^2}{2\pi r}\\I=\sqrt{\frac{2\pi r (F/L)}{\mu_0}}=\sqrt{\frac{2\pi (0.02)(0.225)}{4\pi \cdot 10^{-7}}}=150 A[/tex]
