### Electronic field calculation of 2 wires | Physics Forums

Two nonconducting wires meet at a right angle. One segment carries of charge distributed uniformly along its length, and the other carries distributed uniformly. The problem statement, all variables and given/known data 2 m nonconducting wires meet at a right angle. One segment carries +* Answer to 2Two m nonconducting wires meet at a right angle. One segment carries + µC of charge distributed uniformly along its length, and the other.

And you can probably guess that, but let's go through the exercise. Because I realize that last time I did it, I got a little bit messy. And I'll do it a little bit cleaner this-- I don't have to draw as many magnetic field lines. So let's say that's wire 1. And I'll just make the currents go in opposite directions.

So this is I1. And this is I2. So what would the magnetic field created by current 1 look like? Well, let's do the wrap around rule.

## Statement of a problem № 40673

Put our thumb in the direction of the current, and then the magnetic field will wrap around. It'll go into the page here and it'll go out of the page here. If you put your thumb up like that. Your right hand, always use your right hand. And then you'll get that type of magnetic field.

And of course, it's going into the page, into the video screen, all the way out to infinity. It gets weaker and weaker. It's inversely proportional to the radius away from the wire, so it'll get weaker and weaker. But even here, this magnetic field is going into the page. Now we know, just as a little bit of review, the force created by current 1 on current that's just the convention I'm using, you wouldn't always put the 1 first-- is equal to current 2 times some length-- let's call that length along the wire.

This is going to be a vector because it's a magnitude of length and a direction. And it goes in the same direction as the current. So let's say that that is L2.

So we're talking about from here to here. Cross product that with the magnetic field. I'll switch back to that. The magnetic field created by 1.

Now it all seems pretty complicated, but you can just take your right hand rule and figure out the direction. So we put our index finger-- I'm doing it right now, you can't see it-- you put your index finger in the direction of L2. You can write the 2 down here, instead of writing a big 2 up there. Put your index finger in the direction of L2. I keep redoing it just to make sure I'm drawing it right. Put your middle finger in the direction of-- so this is L2.

Goes in the direction of the middle finger. Sorry, the index finger. Your middle finger is going to go in the direction of the field. So it's going to be pointing downwards, because the field is going into the page, on this side of this wire.

And then your other hands are going to do what they will. And then your thumb is going to go in the direction of the net force. So your thumb is going to go like that.

So there you have it. This is the top of your hand. You have your little veins or tendons, whatever those are, that's your nail.

## Two 1.20m nonconducting wires meet at a right angle.?

So in this situation, when the current is going in opposite direction, the net force is actually going to be outward on this wire. The net force is outward. And then if you don't believe me, you might want to try it yourself, but the force on current 1 or on wire 1, or some length of wire 1, caused by the magnetic field due to current 2, is also going to be outwards.

So here, if you want to think about it little bit, or have a little bit of intuition, if the current's going in the same direction they will attract, and if currents are going in opposite directions they will repel each other.

So anyway, let's apply some numbers. Let's apply some numbers to this problem. Let's do it with the opposite current direction.

So let's say that current I'm just going to make up some numbers-- is 2 amperes. Current 2 is, I don't know, 3 amperes. What else do you need to know? We need to know how far apart they are. So let's say that this distance right here is, I don't know, let's say it's small.

Let's try to get a respectable number. Let's say that they're 1 millimeter apart.

**The Electric Field Due to a Line of Charge**

But we want everything in our standard unit, so that all the units work out. So let's convert it to meters.

### Physics 21 Solutions

So that equals 1 times 10 to the minus 3 meters. So they're pretty close apart. Now let's figure out the-- well, let's do the force on wire 1 due to current 2. Just so that we can see that this is also repelled. So let's say that the length in question, L1, is equal to-- I don't know, let's make it a long wire-- 10 meters. So how do we do this?

So first let's figure out the magnitude of the magnetic field created by I2. I drew this hand too big, took up too much space. So the magnetic field created by current 2, worried about the magnitude of it, that is equal And we saw before, we're assuming that these are-- it's in air.

So we can use the permeability of a vacuum. So it's equal to that constant, the permeability of a vacuum. Now remember we figure out the direction by wrapping our hand around it. It can be seen that at the position of charge q3the electric field points down and to the right.

### Magnetic force between two currents going in opposite directions (video) | Khan Academy

This is in agreement with the calculations we just did. We will apply the full vector version of the Coulomb force law to find the force exerted on q3 by q2 F2 on 3then the force exerted on q3 by q1 F1 on 3then apply the superposition theorem to find the total force on q3 F3. We are now ready to work out the force exerted on q3 by q2: We will use the Pythagorean theorem to calculate the magnitude of a vector from its components.

To answer questions c and dwe need to find the magnitude and direction of this vector. Note that part d on MasteringPhysics expects an answer in radians.

A uniform horizontal electric field exists in this room. Find the x- and y-components of the net electric field at the origin. Remember that we must do a separate integral for each component E. One segment carries 2. Find a the magnitude and b the direction of the electric field these wires produce at point Pwhich is