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

  • Thread starter Thread starter tandoorichicken
  • Start date Start date
  • Tags Tags
    B-field Wires
AI Thread Summary
When one parallel wire carries a current of 100A, the current in the other wire can be calculated using the force per length equation for parallel wires. Given a force per length of 1.00*10^-3 N/m and a separation of 2.50 cm, the current in the second wire is determined to be 0.0005 A, or 500 mA. This indicates that the second wire carries a significantly smaller current compared to the first. The calculations illustrate the repulsive force between the wires due to their currents. Overall, this scenario exemplifies the principles of magnetic field interactions between parallel conductors.
tandoorichicken
Messages
245
Reaction score
0
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.
 
Physics news on Phys.org
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? :smile:
 


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.
 
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
Back
Top