What is the current in the other parallel wire when one carries 100A?

[tandoorichicken]

Two parallel wires repel each other with a force per length of 1.00*10^-3 N/m when spaced a distance of 2.50 cm apart. If one wire has a current of 100A, what is the current in the other wire?

Don't know where to start this one.

 

[Chen]

Two equations. First, the magnetic field that a wire with current creates at a distance of $$r$$ from it is:
$$B = \frac{\mu _0I}{2\pi r}$$
Second, the magnetic force that operates on a wire with current inside a magnetic field is: (When the magnetic field is perpendicular to the direction of the current)
$$F_m = IBl$$
To find the force per length, just divide it by $$l$$.

Now can you solve it?

 

[M98Ranger]

To calculate the current in the other wire, we can use the equation for the force between two parallel wires:

F = μ0*I1*I2*ℓ/2π*d

Where F is the force per length, μ0 is the permeability of free space (4π*10^-7 N/A^2), I1 and I2 are the currents in the two wires, ℓ is the length of the wires, and d is the distance between them.

In this case, we are given the force per length (F), the distance between the wires (d), and the current in one of the wires (I1). We can rearrange the equation to solve for I2:

I2 = 2π*d*F/(μ0*I1*ℓ)

Plugging in the given values, we get:

I2 = 2π*(0.025 m)*(1.00*10^-3 N/m)/(4π*10^-7 N/A^2*100A*1m)

Simplifying, we get:

I2 = 0.0005 A

Therefore, the current in the other wire must be 0.0005 A, or 500 mA. This shows that the current in the other wire is much smaller than the current in the first wire, which makes sense as the force per length is also much smaller. This demonstrates the principle of B-field repulsion, where two parallel wires with currents flowing in the same direction will repel each other.