# One Question about Long wires and Mag fields, want to verify please.

1. Jul 28, 2004

### Sam_The_Great

Hello, I have this problem and I want to verify if somebody thinks I did it right.

The figure shown below shows an infinitely long wire carries current I1 = 20A. A short wire of length 9cm lies perpendicular to the direction of the long wire. The nearest end of the short wire is 3 cm away and it carries a current I2 - 5A in the direction show. Thwat is the net force on the short wire?( The magnetic field produced by the infinitely long wire is not uniform)

I = 20 A

------------>---------------->----------------------->----------------
|
| 3cm
|
|
|
| 9cm
down current
|
| I = 5 A

The approach I took was that I said F = Ilb and we know that B = u02I1/4piR
R in this case is y, we know that the change in force dF = I2u02I1/4piy dy and you integrate from .03 to .12. The answer I get is 3.58 N in the x direction. Would this be the right approach?

2. Jul 28, 2004

### maverick280857

The approach looks okay....do you have a problem understanding why you did this or if this is correct?

To summarize, you consider an element dy of the wire carrying current I2 at a distance y from the source of the field (the wire carring current I1). You find the expression for the force dF between the current setting up the source of the field and a current element I2*dy. Finally you integrate over the length of the wire to get F(total).

Cheers
Vivek

3. Jul 28, 2004

### Sam_The_Great

Thanks vivek. Another part of the question, how would it change that the infinite wire's magnetic field is not uniform versus being uniform.

From my equation F = Ilb, do you think I took the right approach to get the dF equation? I don't completely understand that equation, but I somewhat took the parallel current line equation and went off of there.

Thanks again

4. Jul 29, 2004

### maverick280857

Okay Sam, think of this as follows: a field is just a mathematical function. If it is uniform in a region of space, it has a constant value in that region, which is independent of where you place a point of observation in the region. If it is non-uniform, it means that the field is a point function...that is, its value depends on the place where you measure it.

Mathematically, if the infinite wire's magnetic field (lets call this B(source)) is non uniform, you cannot (always) take B out of the integral,

$$F = \int_{y_{i}}^{y_{f}}IBdy$$

(where $$y_{i}$$ and $$y_{f}$$ are suitably chosen limits of integration for the variable y, representing the end points of the finite wire).

More than that though it means that B is now dependent on y (that is if B is a function of y and it varies radially) and so successive current elements will be subject to different Lorentz forces (though the deceptive expression dF = IBdy makes you think otherwise, except if you write here the expression for B and substitute it into the expression for dF). To see how, you could integrate from y = 0 to some y and get a function for the force which depends on y. Now compare it with the function that you would get if you pulled B out of the integral (if B were uniform).

Does that answer your question?

As far as your approach is concerned, it is correct. Perhaps the trouble you are facing with it is that you have been exposed to a problem involving Lorentz Forces and Integration to get to the final answer. The idea of the infinitesimal element being considered for forces first and then being integrated to get the correct expression comes from the study of forces in classical mechanics, which you have used here. You should be very familiar with this idea since it is used quite frequently in electrodynamics especially in a topic called Electromagnetic Induction where you will face similar problems except that now, there will be a relative motion of the source (field) object and the effected object (in order to change a quantity called the magnetic flux).

Cheers
Vivek

Last edited: Jul 29, 2004
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