### Video Transcript

In the figure, π, π, and π are three long, current-carrying wires. Which of these wires is not affected by a magnetic force?

Okay, so in this question, we can see that first of all weβve got a wire π. And within this wire, thereβs a current πΌ moving in the upward direction, as has been drawn in the diagram. Now, as well as this, we can see that thereβs a wire π which we assumed to be parallel to wire π. And wire π has a current three πΌ also in the upward direction. Finally, there is wire π which is also parallel to π and π. And thereβs a current two πΌ in the upward direction once again.

Now, as well as the fact that we have three long current carrying wires that are parallel to each other, we also know that the distance between wire π and wire π is π. And the distance between π and π is two π. Now, weβve been asked to find which of these three wires is not affected by a magnetic force. In other words, which of the wires does not experience a magnetic force? Now, for a wire to not experience a magnetic force, it needs to be in a region where the magnetic field is zero because if there is a magnetic field at the location of a wire that the wire will experience a magnetic force due to that magnetic field.

So weβve established that we need to choose one of the three wires. And that wire will be in a region where the magnetic field is zero. But then where are these magnetic fields even coming from? The answer to that is the other wires. Itβs worth remembering that a long straight current-carrying wire will produce a circular magnetic field around it at every point along the wire. The direction in which the magnetic field points is given by Ampereβs right-hand rule.

Ampereβs right-hand rule tells us that if the thumb on the right hand point to the direction of the current along the wire, then the direction in which the fingers curl gives us the direction of the magnetic field. Therefore, we can see that, in this situation, the magnetic field points in this direction. And so using Ampereβs right-hand rule, we can see that each of the wires in the question are going to produce their own magnetic field. Now, of course, weβve only chosen to draw a certain number of circles but this magnetic field extend to very large distances. And this is true for all of the fields from all of the wires.

Interestingly, the magnetic fields get weaker the further away we get from the wire. And this is going to factor into the fact that one wire is gonna be at location where the magnetic field overall is zero. So with that in mind, letβs consider the magnetic field at wire π. Now, the magnetic field in which the wire π is going to be present is going to be a result of the magnetic field produced by the wire π and the wire π. And so weβre gonna have to add up the contributions from wire π and wire π.

But before we even get into the mathematics of it, letβs look at it a bit more from a geometrical perspective because we know what direction the magnetic field due to wire π and wire π are pointing in. And so we can use this directions to work out whether the overall magnetic field at π could even be zero. What we mean by this is the following. Letβs imagine the magnetic field from π or at least a particular magnetic field line from π that crosses the wire π. Now, we havenβt drawn all of the little circles inside this. But we donβt need to. Weβre only considering the magnetic field due to π at πβs location. And we know that the magnetic field due to π points in this direction. And so at πβs location is going to be pointing in this direction or more specifically out of the screen, in other words, pointing towards us watching this whole setup.

Now, we can do exactly the same thing for the magnetic field from π at the location of π. Now, hereβs the magnetic field line from πβs magnetic field that intersects the wire π. Once again, we know that itβs a circle. And it is pointing in this direction. And so at the location of π, the magnetic field from π will also be pointing in this direction, that is, out of the screen towards us. And the same direction as the magnetic field from the wire π. So even the wire π is really far away from wire π. And so the strength of the magnetic field due to π at πβs location is going to be weak. The fact of the matter is that the magnetic field due to π and the magnetic field due to π at πβs location are both pointing in the same direction.

So no matter how weak or strong they are, they can never cancel out. And so we can never have a zero magnetic field at π βs location. Therefore, no matter how weak or strong, wire π will experience a magnetic force in the setup that weβve been given in the question. So we can immediately rule out wire π as an answer to our question. And having done that, we can apply the exact same logic to wire π because firstly we can consider the magnetic field line from π that intersects the wire π. And we can see that it is going to be pointing in this direction once again. And just as a quick reminder thatβs been given to us by Ampereβs right-hand rule. And so at πβs location, the magnetic field is going to be pointing in this direction, into the screen or away from us, the observers.

And similarly the magnetic field line from π that intersects π is going to be pointing in this direction and so, at the location of π, is going to be pointing into the screen once again. Therefore, both the magnetic field lines from π and from π that intersects π are pointing in the same direction and therefore can never cancel out. So once again, we know that the magnetic field at π can never be zero even though they could be potentially very weak there. But we donβt care about whether itβs weak or strong. We just care if itβs zero. And in this case, itβs not.

So now that weβve ruled out wires π and π as the answer to our question, letβs actually confirm that wire π is the correct answer. We can see that the magnetic field line from π that intersects π is pointing in this direction and, therefore, at πβs location, is pointing into the screen. However, the field line from π at πβs location is pointing in this direction and so, at πβs location, is pointing towards us. Thatβs out of the screen. So it is possible that the two magnetic fields, the one from π and the one from π, cancel out at πβs location.

So letβs make sure that that does happen. Now we might think that things might not work out the way we expect because π is actually a lot closer to π than π is. The distance between π and π is π. But the distance between π and π is two π. However, itβs also important to remember that the current in π is twice as strong as the current in π. So even though π is closer, the current in π is stronger. Now, itβs at this point that we can recall that the magnetic field produced by a long straight current carrying wire is given by multiplying π, thatβs the permeability of the space that the wires are in, by πΌ subscript π, thatβs the current in the wire, and dividing this by two π and the distance between the wire and the point at which weβre calculating the magnetic field.

In other words, if this was a long straight current-carrying wire and we were trying to calculate the strength of the magnetic field at a point that was π away from the wire, then we will find that magnetic field strength to be π multiplied by the current in the wire divided by two π and π. So with this logic, we can work out the total magnetic field at this point here which is where the two magnetic fields due to π and π interact. And importantly this is at πβs location. So we can say that the total magnetic field at πβs location, which we will call π΅ subscript π, is given by adding the contributions both from πβs magnetic field and πβs magnetic field.

And we also need to remember that theyβre pointing in opposite direction. So one of them is going to have a negative sign. It does really matter which we can choose. So letβs say that the contribution due to πβs magnetic field is given by π multiplied by the current in wire π, thatβs πΌ β β so weβre multiplying π by πΌ β β and then divided by two π and the distance between π and the point at which weβre trying to work out the magnetic field. So thatβs this distance here. But then that distance is π. So we multiply two π by π. And then to this contribution, weβre going to subtract πβs contribution because, remember, we said one of them needs to be negative because theyβre pointing in opposite directions.

So weβve arbitrarily said that the contribution from πβs magnetic field is positive because itβs pointing into the screen whereas πβs contribution is negative because itβs pointing out of the screen. We could have easily done it the other way around. It wouldnβt really matter. But anyway, so πβs contribution is π multiplied by the current in wire π, which is two πΌ this time. And so we multiply π by two πΌ. And we divide this by two π multiplied by the distance between π and the point at which weβre calculating the magnetic field. So thatβs the distance between here and here which is two π.

So we put two π in the denominator. Now at this point, we can see that in this fraction over here, weβve got a factor of two that cancels out. And so the fraction simplifies to ππΌ divided by two ππ. So the total magnetic field at πβs location is given by ππΌ divided by two ππ minus ππΌ divided by two ππ. In other words, this magnetic field is zero just as we needed it to be. And so weβve come to the conclusion that the magnetic field at πβs location is zero. And therefore wire π is not affected by a magnetic force. And that is the answer to our question.