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**Example 1- Parallel Wires**

Let’s do an example related to the parallel current carrying wires. Let’s assume that we have two parallel wires and from the top view, both of them are carrying a current into the plane direction. Let’s say the first wire’s current is *i**a*. The other one is *i**b*. And these two wires are separated from one another by a distance of *d*. We’d like to find a point at which the net magnetic field is 0. Find a point such that **B****total** at that point is equal to 0.

Okay. First, let’s try to determine the direction of the magnetic field in different regions. If we look at points on the left-hand side of wire *a*, applying right-hand rule, keep pointing to the right-hand thumb into the plane, and circling the right-hand fingers about the thumb, we will see that the magnetic field lines associated with current *a* will be rotating in clockwise direction. Therefore on the left-hand side, the magnetic field is going to be tangent to the field line passing through the point of interest. So at this point, it is going to be pointing in upward direction along this line.

Similarly, for *i**b* also, applying the right hand rule, we will see that the field lines are going to be circling in clockwise direction. And the one which is passing through at a point which is on the left-hand side of wire *a* will also be pointing in upward direction. Since the magnetic fields generated by current *a* and current *b* at all points on the left hand side of this wire will be pointing in upward direction, they will never cancel one another and we will never find a point that the net magnetic field will be 0.

On the other hand, if you consider the right-hand side region of *i**b* along the line joining these two wires, and in this case we will see that by applying the right-hand rule, we know that the field lines are circling in clockwise direction for both of them, so *i**b* will generate a magnetic field in downward direction, tangent to the field line, and as well as *i**a*. Therefore, on the right-hand side of current *b*, also the magnetic fields generated by these two currents will add, and we’re going to end up again, with no point that will have total magnetic field is equal to 0.

So we’re left the region between these two wires. If you consider the points over here, let’s say some that are over here, for example, the magnetic field lines due to current *a* circling in clockwise direction at the location of this point. Therefore, the magnetic field due to current *a* will be pointing in downward direction, **B****a**. And for the current *b*, again, they’re circling in clockwise direction at the location of this point. The magnetic field associated with current *b* will be pointing in upward direction. Therefore, this region is the region that we’re looking for, because in these regions, magnetic field lines are going to be aligning in opposite directions due to these two current *i**a* and *i**b*.

Well, whenever they’re equal in magnitude, the we will have the point what we’re trying to find. Let’s assume that at this point, the magnitude of these two magnetic field vectors are equal to one another. Let’s denote this distance as *x*. Therefore, this distance is going to be equal to *d* minus *x* relative to current *b*.

Well, we know that magnitude-wise *B**a* is equal to *μ**0* *i**a* over 2*π* times the distance between the wire and the point of interest, and that is what we called *x*. And *B**b* is equal to *μ**0* times *i**b* over 2*π*, the distance between the wire and the point of interest. Let’s say that now that point is *P*, and that is *d* minus *x*.

We want these two magnetic field vectors to be equal to one another so that **B****total** is going to be equal to vector sum of **B****a** and **B****b**. And since they are in opposite directions, whenever they become equal, when we add them **B****total** is going to be equal to 0.

Therefore if we equate these two equations, *μ**0* times *i**a* over 2*π* *x* should be equal to *μ**0* times *i**b *over 2*π* *d* minus *x*. Here we can cancel *μ**0*’s from both sides and as well as 2*π*‘s. That leaves us an expression such that *i**a* times *d* minus *x* becomes equal to *i**b* times *x*.

We will rearrange this expression for *x*. If we open the parentheses on the left-hand side, we will have *i**a* times *d* minus *i**a* times *x* is equal to *i**b* times *x* on the right-hand side. And collecting *x* terms on one side, *i**a* *d* will be equal to *i**a* plus *i**b* times *x*. And solving for *x*, we’re going to end up with *x* is equal to *i**a* *d* over *i**a* plus *i**b*.

Therefore, from the wire *a*, whenever we’re this much of distance away, then the net magnetic field at that point is going to be equal to 0.