. Two long, parallel wires are separated by a distance of 0.400 m (Fig. E28.29). The currents I1 and I2 have the directions shown. (b) Each current is doubled, so that I1 becomes 10.0 A and I2 becomes 4.00 A. Now what is the magnitude of the force that each wire exerts on a 1.20-m length of the other?

Video transcript

Welcome back everybody. We are taking a look at two infinitely long parallel conductors which I'm going to represent with these two parallel arrows on both ends here we are told that they are separated by a distance of our And we are actually going to make a closer observation at a shared length of L between the two conductors here. Now we are told that they exert a certain force on each other at the current given conditions. But we are tasked with finding, say we triple the current. What will be the strength of force exerted on each conductor by one another? Well, as it stands with the current conditions, we have the strength of force according to this formula. Right here we have new not times our initial current squared times R length divided by two pi R. Now here's the thing. We are going to plug in a new current. That is triple the old current. So let's go ahead and plug this in into our equation. We then get our equation is new, not times three times our initial current squared times L. All over two pi R. What I can do is I can square this three and then take it out and put it in the front. So we then have three squared is nine. So there will be three times mu not times I. R. Finish current squared times L. All over two pi R. But what is this value? Right here? Well, that is just our original strength of our force. Our original magnitude of force. So the new magnitude force will be nine times the magnitude of the initial force corresponding to our answer choice of C. Thank you all so much for watching hope this video helped. We will see you all in the next one.