# Homework Help: Calculating Force in Parallel Conductors

1. Aug 29, 2004

### voormann

This question actually has nothing to do with homework, since I am not a student, it's simply to satisfy my need for understanding the relationship between force and parallel conductor dynamics..

The ampere is that constant current which, if maintained in two straight parallel conductors of infinite length, of negligible circular cross-section, and placed 1 m apart in vacuum, would produce between these conductors a force equal to 2 x 10^–7 newton per metre of length.

I would like to know the equation used to calculate the current and force between 2 straight parallel conductors of infinite length and negligible cross-section, per given length, at a given distance apart in a vacuum.

first I would like to know what precise current in Amperes is required to effect a force of 89,2654465915787 Newtons per 0,84729456 metre of length, if we plug in a new distance between conductors: 0,84729456 metre

thanks!

Last edited: Aug 29, 2004
2. Aug 30, 2004

### Tide

The force per unit length on the wires varies as the product of the electrical currents and inversely with their separation. Given the definition of the ampere and using ratios and proportions you should now be able to answer your questions.

3. Aug 30, 2004

### voormann

thanks!

so the force increases in direct proportion to the increase in current
and decreases as seperation increases, is this correct?
is the increase in force with seperation linear or logarhithmic?

4. Aug 30, 2004

### Tide

The magnitude of the force is $$F = k \times \frac {I_1 I_2}{d}$$ where d is the separation.

5. Aug 31, 2004

### maverick280857

Just to add to Tide's post, we have

$$F = k\frac{I_{1}I_{2}}{d}$$

where d is the distance and I1 and I2 are the currents carried by the wires. If you have read about Lorentz Force and Magnetic Fields (the B vector and F = ILB where L = length of conductor), you can actually derive the relationship for the force between two parallel straight (long) current carrying conductors. Try it out to get an insight.

Cheers
Vivek.