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**Example- Magnetic field of a current loop**

Let’s do another example related to the application of Biot-Savart Law, and in this case let’s try to calculate the magnetic field of a circular current loop along its axis.

Magnetic field of a circular current loop along its axis:

In this case let’s consider a circular wire, something like this, and carrying a current in, let’s say, counterclockwise direction, and let’s say the radius of this wire is R and we’re interested with the magnetic field that this current loop generates along the axis of this circular wire, let’s say, some Z distance away from its center at this point. In order to calculate this field we apply Biot-Savart Law and it’s given as Mu zero over 4 Pi, integral of i dl cross r over r cube. Here i dl is an incremental current element the we chose in the direction of flow current therefore if we choose such a current element, let’s choose something right over here and that current element is going to generate a magnetic field at the point of interest, position vector r is a vector drawn from this element to the point of interest therefore it’s a vector something like, and again this distance is, naturally, equal to big R, the radius of this ring, and so this incremental current element will generate an associated incremental magnetic field at this point. We’ll determine the direction of that magnetic field from this cross product over here and also therefore moving the right-hand fingers in the direction of the first vector, which is i dl in this case. If you visualize this picture from the cross sectional point of view then we will see that this current is coming out of plane at this point and flowing through the wire and going into the plane at this point, so if i dl is coming out of the plane here and r is this way, then after holding the right-hand fingers in the direction of i dl out of plane first and adjusting them such that we can curl towards the second vector, which is on the plane then we will end up with and magnetic field vector, which will be perpendicular to all of these vectors. And, again, also you can visualize that the magnetic field line generated by this i dl on this plane is going to in the form of a circle with a radius of little r here and the magnetic field will be tangent to that field line at the point of interest. Since current is coming out of plane and holding the right-hand thumb in the direction of current out of plane, and then curling right-hand fingers about the thumb will give us the field line passing through this point will therefore be in the form a circle and rotation in counterclockwise direction, therefore the magnetic field will be tangent to that field line and is going to be pointing, therefore, in this direction d B. So at that point it is tangent to the field line, which is a circle going in counterclockwise direction and the magnetic field is tangent to that field line passing through this point of interest P.

Well, when we look at the geometry of our current we see that it is naturally symmetric relative to the axis over here, therefore we can also find another incremental current element, which is going to be right across from this one and that one will be going into the plane from the cross sectional point of view, and the associated position vector will be drawn from here to the point of interest, something like this and if we look at the magnetic field that this incremental current element generates at this point and by apply Right-hand Rule it will generate a magnetic field, i dl cross B which will be pointing such that perpendicular to this position vector r and therefore it’s going to be something like this. And, again, here we can apply the Right-hand Rule in the since that holding the right-hand thumb in to the plane and circling or rotating right-hand fingers about the thumb we will see that the magnetic field of this current over here, which it just going into the plane, will be in the form of concentric circles, magnetic field lines, and the one which is passing through this point will be a circle something like this, and in this case it is going to be rotating in clockwise direction so the magnetic field will be tangent to this field line at the point of interest P, which is just right over here.

Now if you do a similar type of analysis for all the incremental current elements along this loop we will see that the associated magnetic field lines are going to be falling along the surface of a cone something like this, so these fields are going to be lying along the surface of the cone. Well, of course the vector sum of all these incremental magnetic field vectors will eventually give us the net magnetic field at this point.

If you just consider these two pair and if you add the vectorially, and in order to do that, of course we first introduce a coordinate system, let’s say, something like this, and resolve these vectors into their components by taking their projections along the horizontal axis, like this and as well as along the vertical axis we will see that the vertical components are going to be in the same direction, but the horizontal components d Bh, let’s denote that with d B’s of h are going to be aligning in opposite directions.

Well, since i dl’s will have the same magnitude and these source currents are same distance away to the point of interest along the axis of this current loop, then the magnitudes of dB’s generated or the incremental magnetic field generated by these increment al currents will be the same. It also means that their components will have the same magnitude and as a result of this the horizontal components, since they are aligning along the axis in opposite directions with the same magnitudes, when we add them vectorially they will cancel. Therefore we can make a note by saying that dB horizontals cancel due to the symmetry. Therefore, what we’re left are just the vertical components of all these incremental magnetic field vectors and if we add them then we will end up with the total magnetic fields along the axis of this current loop. The addition process over here is the integration therefore the sum of all these vertical components will give us eventually the total magnetic field generated by this current loop.

Okay, let’s try to express these vertical components in explicit form. Of course, in order to do this we will take the advantage of these right triangles formed as we resolve these incremental magnetic field vectors into their components with respect to this coordinate system. Let’s consider this angle, or define this angle and call as Theta. In that case, by considering this right triangle over here and applying the trigonometric identities over here the Cosine of this angle will be equal to the ratio of the adjacent sides to the hypotenuse of this triangle. Therefore, we can express dB vertical as hypotenuse times Cosine of Theta. In other words, the side of the triangle is going to be the side that we’re after, which is the vertical component of the incremental magnetic fields.

Well, in this expression we will calculate the magnitude of the magnetic field dB from the Biot-Savart Law and we need to express Cosine of Theta in terms of the quantities which are given in our problem. If we look at this angle over here, angle Theta, we see that this side of that angle is perpendicular to this side, the little r side, and also the other side of the same angle Theta, which is this one, perpendicular to this side, Big R. Therefore, this angle, angle Theta, and this angle are congruent angles. In other words, they have mutual perpendicular sides so they have to be equal. Then this angle is also Theta.

Now we can use this large triangle, which is forming from the distances, it’s a right triangle with this angle being 90 degrees, and if we express Cosine Theta for that triangle that becomes equal to the ratio of the adjacent side, which is big R, to the hypotenuse, and that is little r. Again, in the same triangle if we apply the Pythagoras theorem little r is the hypotenuse of this triangle, therefore r squared is going to be equal to the sum of the squares of the other two sides, which are Z square, plus big R square. Then we can express Cosine of Theta as little r over square root of Z square plus big R square.

All right, and dB from Biot-Savart Law will be equal to Mu zero over 4 Pi, the magnitude of the incremental magnetic field, the magnitude of the first vector, which is i dl magnitude times the magnitude of our vector and that r, and Sine of the angle between these two vectors, so we’re looking for the angle between i dl and r vector. If we consider any one of these, here i dl vector is going into the plane and here i dl vector is coming out of plane. Whereas, position vector r is on the plane indicating that the angle between these two vectors i dl and position vector r is 90 degrees divided by r squared, r in the numerator will cancel the r square in the denominator, and Sine 90 is just 1, therefore the magnitudes of the incremental magnetic field generated by the incremental current element i dl becomes equal to Mu zero i dl over 4 Pi r.

Okay, the total magnetic field is the sum of all these vertical components and the summation is integration, therefore if we express these in explicit form the magnetic field is going to be equal integral of Mu zero i dl over 4 Pi, for little r we can right is down as square root of–actually this quantity over here is r cube, so after cancellation we will have r square back in the denominator–and the r square is Z square plus big R squared, and this part is therefore is the magnitude of the incremental magnetic field dB, and then we have the Cosine of Theta so we will not apply this quantity by Cosine Theta, which is equal big R over square root of Z squared plus R squared.

All right, if we continue, magnetic field is going to be equal to–when we look at this expression here Mu zero is permeability of free space, which is constant, current is constant. 4 Pi Z is the distance between the origin, or the center, and the point of interest, that is constant, and big R is the radius of the current loop, that is also constant. Therefore, all these quantities here, they’re all constants, we can take it outside the integral. Then magnetic field becomes Mu zero i R divided by 4 Pi, Z squared plus R squared times the square root of Z squared plus R squared, will give us Z squared plus R squared to the power of three and a half. And, inside of the integral we’ll have dl.

Now, the integral is going to be taken over the length of this current loop and if you look at over here the i dl is basically an arc length, I mean dl is an arc length, so it subtends therefore a certain angle and if we look at that angle it’s going to be something like this. Let’s denote that angle as d Phi. So, if I redraw that segment over here, in large picture it will look like something like this. This is the i dl element and it subtends a certain angle of d Phi and the radius over here, the radius of the arc, is big R. As you recall by using the definition of radian then we can express this arc length dl as, let’s just consider the distance part of it dl, as radius times the angle that it subtends, which is d Phi. Therefore, the magnetic field becomes equal to M zero i R over 4 Pi, Z squared plus R squared to the power of three and a half, integral of dl, which is R d Phi. We will integrate this quantity throughout the loop and as we do that we will see the corresponding Phi is going to start from zero and will go all the way around to 2 Pi radians to complete the loop, therefore the boundaries will go from zero to 2 Pi radians.

Magnetic field then will be equal to–here again, R is constant, we can take it outside of the integral, which will give us Mu zero i R square in the numerator and in the denominator we will have 4 Pi, Z squared plus R squared to the power of three and a half. The integral of d Phi will give us Phi and if we substitute the boundaries it is going to give us 2 Pi for the first one and zero for the second boundary. Finally, therefore, B is going to be equal to M zero i, R squared over 4 Pi, Z squared plus big R squared to the power of three and a half, times 2 Pi. We can cancel 2 and 4 over here leaving us 2 in the denominator and as well as cancel the Pi’s.

So, the magnetic field therefore become equal M zero i, R squared over 2, Z squared plus big R square to the power of three and a half. We can express this in vector form if we define or introduce a coordinate system of x, y, and z and recalling the unit vectors along these directions as i, j, and k. Since the magnetic field will be in a vertically outward direction the associated unit vector in that direction is k, therefore we can multiply this by the unit vector pointing in outward direction and express the resultant magnetic field in vector form like this.

Here, if we write this expression by multiplying numerator and denominator Pi we will have Mu zero i Pi R squared over 2 Pi, Z squared plus R squared to the power of three and a half. This Pi R squared is going to give us the area surrounded by this current loop. So here, A is the area surrounded by the current loop. In terms of this quantity magnetic field can be expressed as Mu zero times i times A divided by 2 Pi, Z squared plus R squared to the power of three and a half.

Well, when we look at our current loop it is carrying current i and also is surrounding a certain area of A. These two quantities are unique features of this current loop, the area that it surrounds as well as the current which flows along this loop. Therefore, the product of these two quantities will also be a unique feature of this loop.

Furthermore, if we just visualize instead of having one loop we have, let’s say, n turns or n loops as the current goes around this loop it goes along these turns. In that case the magnetic field generated by each loop will add to the next one. In other words, if I had n number of turns then the total magnetic field will n times the magnetic field generated by a single turn. Let’s say then, for n number of turns–that is also the equivalent of saying loops–the magnetic field is going to be equal to, B is going to be equal to M zero over 2 Pi, times N times i times A, divided by Z squared plus big R squared to the power of three and a half.

Now, when we consider the N number of turns, that the number of turns, the current that they carry, and as well as the area that they surround will be unique properties of that specific coil. So, the product of these three quantities therefore will also be a unique property of that coil, of those turns, or those loops, current loops. We have a special name for this product, which we’re going to denote this by Mu and we will call this quantity as magnetic dipole moment vector magnitude.

All right, so okay. Mu therefore is going to be equal, by definition, as N times i times A; the number of turns of the coil times the current flowing through the coil times the area surrounded by the coil. As we can see here N is a scalar quantity, current is a scalar quantity, but A is a vector quantity, so this magnetic field vector, or a magnetic dipole moment vector of this current loop will be in the same direction with area vector, surface area vector that the current loop surrounds.

Now when we were defining the direction of the area vector we said that it always perpendicular to the surface and once the surface is open, in other words, it does not occupy a certain volume then the direction of this area vector can be in either direction as long as it is perpendicular the surface. For example, when we consider a surface surrounded by a circle like this we said that it can be either up or down, both of these are okay, and the magnitude is directly proportional, or equal, to the surface area surrounded by that circle.

Well, when we have the surfaces which are forming or surrounded by current loops in the case we don’t have this freedom to assign a direction of the area vector, each one of these directions. As a matter of fact, for this case we have to follow Right-Hand Rule. If you make a note of this over here by saying that direction of surface area vector for a surface surrounded by a current loop is such that: First, hold your right hand fingers in the direction of flow of current and curl them along the current loop. As you curl your right-hand fingers about the current loop in the direction of flow of current, your right-hand thumb in open position, or let’s say up position, gives the direction of the area vector associated with the surface surrounded by the current loop.

Therefore, if I have a current carrying loop something like this, and if the current is flowing along this loop in counterclockwise direction, then the associated surface areas vector of this surface, the surface surrounded by this current loop, is going to be, holding your right-hand fingers in the direction of flow of current and curling them along the loop, and keeping the right-hand thumb in up position it will show us that the A is going to be, this area vector for in this case, is going to be pointing in upward direction relative to this surface surrounded by the current.

In a similar way it the current is flowing in clockwise direction along this loop then the associated area vector is going to be in downward direction, like this, because keeping the right-hand fingers in the direction of flow of current and curling them along the loop, our right-hand thumb will point in downward direction for this surface. And, we all know that this area vector is always perpendicular to the surface.

From the definition of magnetic dipole moment vector–which is N the number of turns, times current times the area vector–Mu is in the same direction as this surface area vector therefore for a current loop, something like this, as the current is going in counterclockwise direction the magnetic dipole moment vector of this current loop is going to be pointing in the same direction with the area vector, therefore, it’s going to be something like this. For a circular loop such that the current is flowing in clockwise direction then the area vector is pointing downward direction and the associated magnetic dipole moment vector will also be pointing in downward direction.

Of course these are the magnetic field vectors along the axis of this current loop and if you do a similar type of analysis and look at how the magnetic field lines are distributed throughout the region due to this current loop we will see that they are going to be aligning something like this. If I try to draw several of them, such a current loop is going to generate magnetic field lines of this form . . . and so on and so forth, along the axis the magnetic fields will be tangent to the field line passing through that point and it is going to be in the upward direction. If the current is flowing counterclockwise direction then these field lines are going to be oriented this way. Again, the related magnetic dipole moment vector for such a system will be in the same direction with the area vector associated with this surface or the area surrounded by the current loop.

After we introduce the magnetic dipole moments of a current loop which is pointing in an upward direction, for circular current loop carrying a current in counterclockwise direction, like in this diagram, or pointing in downward direction, like this, for a circular current loop carrying a current in clockwise direction.

Let’s look at a special case. The magnetic field magnitude of such a current loop that we found by applying Biot-Savart Law is Mu zero over 2 Pi [in its place 44:45] N i A–the number of turns times the current and the magnitude of the area surrounded by the loop with the magnitude of the magnetic dipole moment vector, divided by in the denominator we had Z squared plus R squared to the power of three and a half. And here, these are common notations that are used in most of the textbooks, so be careful not to confuse the magnetic dipole moment vector Mu with the permeability of free space Mu zero.

All right, well this was the case that–here’s our current loop and we calculated the magnetic field some z distance away from the center of the loop, which has a radius of R and carrying a current in counterclockwise direction. We found that the magnetic fields at this location is pointing in upward direction and the magnetic dipole moment vector associated with the current loop is pointing also in upward direction. We call the location of this point of interest, relative to the origin or the center, as Z distance away from the center.

Well, now let’s look at a special case such that our point of interest is a far distance away from the loop, on the axis, such that Z is much, much greater than the radius of the current loop. In that case when Z is much greater than R, R over Z will be much, much smaller than 1, this will enable us to make certain approximation for the magnitude of the magnetic field. If we take the quantity in order to have this ratio, if we take Z outside of this bracket we’re going to have Z squared will come out to the power bracket of 3 over 2 as Z cube and inside of the bracket we will 1 plus R squared over Z squared, to the power of three and a half. Now since R over Z is much smaller than 1 then R squared over Z squared will be even smaller when we compare it to 1. Therefore, one can easily neglect this quantity in comparent to 1 then the equation reduces to this simplified form by taking advantage of this ratio Mu zero over 2 Pi, times Mu over Z cube. So we can obtain an approximate expression for the condition that at our point of interest is a far distance away relative to the radius of the current loop, then the magnetic field that the current loop will generate along its axis is going to be equal to this quantity, in vector form again, we can multiple this magnetic unit vector k pointing in upward direction.

This current loop also represents a very good model to be able to explain, which I will talk about this a little bit later, the magnetic behavior of matter. We see that from the atomic structure of the matter, from the planetary model of the atom, the electrons are orbiting about the nucleus of the atom. For some of the materials, as they orbit about the nucleus, since the electrons negatively charged particles they represent as they orbit they represent moving charges. Therefore, they generate these tiny little current loops and since some of the atoms these orbiting electrons that they orbit in a unison way by generating these tiny little current loops and associated magnetic dipole moment vectors. As we see, the magnetic dipole moment vectors are in the same direction with the associated magnetic fields and we will see later on that for some cases they can be aligned in the same direction, therefore all these tiny little magnetic field vectors add to one another by generating a net magnetic field. This is going to eventually lead us to magnetic behavior of the matter, how some certain objects can be magnetized whereas how some other types of objects cannot be magnetized. Therefore these tiny little current loops or the magnetic field of a current loop is a very good model in order to explain such a complex physical phenomena.