Instructor: Let’s do another example associated with the application of Biot-Savart law. In this case, let’s consider a wire which has a semicircular region something like this, and a flat part and another semicircular region something like this.
Let’s assume that this is the common center of these semicircular parts, and we can give some dimensions. Let’s say this semicircular part has the radius of A and this semicircular part has the radius of B and a current I is flowing through this wire in clockwise direction, and we’d like to figure out the net magnetic field that this current generates at the central point let’s say O. Magnetic field at point o is the question mark.
In order to do this, we’re going to apply Biot-Savart law for each segment of this law — this loop, for the upper semicircular region, the lower semicircular region, and as well as these straight regions.
We can number these segments. Let’s say this is the segment one, segment two, segment three, and segment four. Biot-Savart law, the mathematical expression is such that B is equal to mu-0 over 4pi integral of idl cross r over r cube where dl is an incremental displacement vector chosen in the direction of flow of current and r is a position vector drawn from that element to the point of interest.
Therefore if we look at the first segment, here idl, let’s say this is the incremental displacement vector, idl is going to be pointing to the left. The position vector associated with this current element will be drawn from here, from the element, to the point of interest. Therefore, it’s going to be a vector like this.
And so when we write down the magnitude of this magnetic field expression, that is equal to mu-0 over 4pi times integral of idl magnitude, r magnitude, times sin of the angle between these vectors.
So, for the first segment, if we call the magnetic field associated with that segment as b1, the magnitude will be — since the angle between idl and r is 180 degrees for this case, therefore we will have sin 180 degrees divided by r cube. But sin 180 degrees is 0. Therefore there will not be any contribution to the magnetic field due to the current flowing along segment one, so b1 will be equal to 0.
In a similar way, we could consider segment three, the position vector associated with the idl. Here for that segment it’s going to be pointing from idl to the point of interest, and there the angle between idl and r will be 0 degree. They’re both pointing in the same direction, so in this case b3 will be equal to mu-0 over 4pi integral of idl magnitude r magnitude times sin of the angle between them, which is 0 degree, divided by r cube, and in this case also since, since sin 0 is 0, b3 will give us just 0.
Okay now, let’s consider loop two. For loop two, idl is this incremental current element, and position vector relative to this point o is going to be a vector drawn from that point to the point of interest, and the angle between these two vectors, as we can see from the diagram, will be 90 degrees, so the magnitude of the magnetic field that this element generates at this point will be from Biot-Savart law — let’s call that one as b2 mu-0 over 4pi idl times r magnitude, and the magnitude of this vector is the radius of the semicircular region, so that is a times sin of the angle between these two vectors, and that is 90 degrees, divided by r cube, and the magnitude of r is a, therefore we will have a cube over here. So here for this one r magnitude is equal to a.
Now how about the direction? To be able to figure out the direction that this current, incremental current, generates at this point, we will use right-hand rule. Holding the right-hand fingers in the direction of first vector, which is idl pointing this way, and then curling them towards the second vector keeping the thumb in open position, we will see that the thumb, right-hand thumb, is going to be pointing into the plane at this point. Therefore the magnetic field that this incremental current element will generate at this location is going to be pointing into the plane, and obviously all these dl’s, idl’s, will generate the magnetic field in the same direction, so b2 is going to be pointing into the plane.
All right, now we know the direction of the magnetic field generated by that segment, and let’s go ahead and calculate the magnitude of that magnetic field. This a and a cube will cancel. We will end up with a square in the denominator. Sin 90 is just 1, and i is common — I mean constant. We can take it outside of the integral.
Therefore b2 becomes equal to mu-0 i over 4pi, and here we have a square in the denominator, which is also constant. We can take it outside of the integral. And inside of the integral, we will have only dl.
And we’re going to take this integral over all the length of this segment, and if we look at over here, this dl is an incremental length. I mean it’s an incremental element and it’s an arc length. So, it subtends a very small angle of let’s say d phi. And then we can express that dl here as the radius times the angle that it subtends, which is d phi.
Then we can express b2 as mu-0 i over 4pi a squared integral — instead of dl we can write down a d phi since we’re adding all these dl’s along the length of this semicircular region and as we do that the angle corresponding to the arc length dl will start from 0 and go all the way to this point that is pi radians. Therefore phi is going to vary from 0 to pi radians.
This a and a square will cancel, and the integral of d phi will give us phi, which will be evaluated at 0 and pi, and that’s going to give us just pi.
So b2 magnitude is going to be equal to mu i over 4pi a times pi. The pi’s in the numerator and denominator will cancel. Therefore b2 is going to be equal to mu i over 4a, and this one is going to be pointing into the plane, so in vector form we can write this down as into the plane.
Now, if we look at the last segment, which is this one, this will be exactly similar to the previous part except the radius is b here. Therefore we will just simply change the radius with b. And furthermore if we look at the direction of the magnetic field that this current generates at this point o and if we plug the position vector for this element, which is going to be pointing again from element to that point and applying the right-hand rule holding right-hand fingers in the direction of i dl and curling them towards the second vector r pointing this way, and if we do that we will see that the right-hand thumb is going to be again pointing into the plane. In other words, for this case also, the magnetic field generated for this lower segment at this location will be pointing into the plane.
So both b2 and b4 is going to be pointing at the center into the plane. These two vectors will be in the same direction. We can see therefore similarly the magnetic field generated from the lower semicircular region will be equal to mu-0 i over 4 times the radius of this semicircle, which is b, and that too will be pointing into the plane.
Since both of these two vectors are pointing in the same direction, the total, the net magnetic field, is going to be sum of these two, mu-0 i over 4 over a plus mu-0 i over — mu-0 i over 4a plus mu-0 i over 4b. And we can write this in common quantities parentheses, which is mu-0 i over 4, and we will have 1/a plus 1/b inside of the parentheses. Therefore b-total is going to be equal to — having a common denominator over here will give us mu-0 i over 4a plus b divided by a times b. This is the net magnetic field at the center of this current configuration, and it is going to be pointing into the plane vector wise.