Example- Magnetic field of an infinite straight current carrying wire
Let’s do an example related to applications of Biot-Savart law. Let’s consider an infinitely long straight current-carrying wire and try to calculate the magnetic field that this current generates at a specific point in space. Magnetic field of an infinitely long, straight current-carrying wire. So for this case, we have a very long, straight wire. Let’s assume that carrying a current, i in upward direction, and we’d like to figure out, some R distance away the magnetic field that it generate at this point P.
And our wire is such that it is going plus infinity in this direction and minus infinity in this direction. And to be able to do this, we will apply Biot-Savart law which simply states that the incremental magnetic field generated by an incremental segment of a current-carrying wire is equal to permeability of free space µ0 times 4π, times incremental current element of idl x position vector r, divided by the cube of the magnitude of this position vector. And this is the mathematic form of experimental Biot-Savart law.
Well, idl is a current element vector that we choose at an arbitrary location in the direction of flow of current. This is idl. r is the position vector, drawn from this element to the point of interest. Therefore, it will be a vector, something like this, and the magnitude will be equal to µ0 over 4π, times idl vector magnitude, times r magnitude, times sine of the angle between these two vectors, from cross product. And that vector is, I mean, that angle is this angle θ, divided by magnitude of the cube of the magnitude of the position vector r, r3.
We can cancel this r with this r3. And, then, the dB becomes equal to µ0 over 4π idl times sinθ over r2. Okay. When we look at our diagram, we will see that, since this is infinitely long, we can define the origin as this point, and we will always find an incremental current element vector below the origin symmetric to this one over here. So in other words, the same distance below the origin, somewhere around here, we will have, we can find another idl.
And that idl will generate its own position. Maybe I should draw this a little bit above [sounds like], just to be symmetric with this one. Its position vector is going to go over here, and then, the corresponding symmetric idl will be somewhere over here. And, in order to determine the magnetic field that these two incremental current elements generate at this point, we will apply right-hand rule. The right-hand rule says that we will hold the right-hand fingers parallel, or in the direction of first vector.
Let me make a note of that over here. Let’s say, review of right-hand rule. And, as you recall that, if we consider a cross product such that three vectors, A, B, C, let A, B and C, be three vectors, then, or let’s say, such that A is equal to B cross C. To determine the direction of A. As a first step, we hold our right-hand fingers, hold your right-hand fingers, in the direction of the first vector. And here, the first vector is B, the second vector is C, in the cross product. And as a second step, adjust your right-hand fingers such that you can curl them from first vector towards the second vector.
So the first vector was B in this expression, the second vector is C. And the third step is, keep your right-hand thumb in open position, in open, or, let’s say, up position, while you are curling your right-hand fingers toward the second vector, and the direction of your right-hand thumb gives the direction of the resultant vector. In this case, vector A. So if we apply this rule to our case, first we will hold our right-hand fingers in the direction of idl, and then we’re going to adjust them such that we can curl right-hand fingers towards the second vector, which is pointing this way.
And as we do that curling process, we keep the right-hand thumb in up, or open position. And in this case, idl is pointing up, r is pointing this direction and this direction. And holding the right-hand fingers in the direction of first vector idl, and then curling them towards the second vector r, the right-hand thumb is going to be pointing into the plane at the location of our point of interest, P. So the vector, magnetic field vector associated with this idl at this point is going to be pointing into the plane, dB.
And here, we’re going to use this symbol into the plane direction, and we will use that symbol to represent r of the plane direction. And you can recall these directions by visualizing the cross represents the tail of an arrow going into the plane, and that represents the tip of an arrow coming out of that plane. Well, if you go back to our diagram, now, we found that the upper dl will generate a magnetic field into the plane direction.
If you go and do this, apply the same process for the lower dl over here, idl, now again, idl is pointing in upward direction. Holding the right-hand fingers in the direction of this vector, and then curling them towards the second vector, r over here, this direction, so curling this way, and holding the thumb in open position or up position, we will see that idl cross r also will generate a magnetic field dB into the plane at the location of our point of interest. So, both of these incremental current elements will generate a magnetic field at this point, pointing in the same direction. In other words, we can conclude, then, the magnetic field generated by the upper half of this current-carrying wire will add to the magnetic field generated by the lower half.
Once we determine this dB as a result of this incremental current element segment, then we go ahead and calculate, or do the same, or apply the same process for the next incremental current element idl, and then for the next one, and then next, for the next one, and so forth. And calculate the associated incremental magnetic field generated at this point. Once we do that, then we add all of them vectorially, to be able to get the total magnetic field.
Of course, the addition process over here is going to be integration. So, the magnetic field will be equal to integral of all these dB’s, which is going to be equal to integral of µ0 over 4πe, and we calculated the magnitude as idl sinθ over r square. And this will be integrated along the whole length of the wire, current-carrying wire, that we’re interested with. Of course, to be able to take this integral, first, we have to put the integrant in an integrable form.
And first, we know that the µ0, 4π and i, these are all constant, we can take it outside of the integral. And second, let’s look at this sine of the angle between idl and r. If this angle is θ, then this angle will be equal to π minus θ. And if I look at sin (π – θ), and open this in explicit form, it will be equal to sinπ sinθ–sorry, sinπ cosθ – cosπ sinθ. Sinπ is 0, cosπ is minus 1. Therefore, this is going to be equal to sinθ.
Well, if I go back to my diagram . . . Let’s say that this, since the wire is located along the vertical axis, let’s introduce a coordinate system, and say that positive y is pointing in upward direction, and, therefore, this idl, relative to the origin, will have a specific y value along this coordinate. And let’s call that distance as therefore y. Now, in this triangle, which is a right triangle, with this angle as being 90 degrees, we can express sinπ – θ, and now we know that it is also equal to sinθ, and that will be equal to the ratio of opposite sides relative to this angle, which is R to the hypotenuse, and that is the r.
And the magnitude of that r, since it’s a hypotenuse in this right triangle, applying Pythagorean Theorem, the r is, or r2 is, y2 plus R2. Therefore, we can express over here, for r, as y2 plus R2 in square root, and sinθ is equal to R divided by r, or R divided by square root of y2 plus R2. Okay. Now, by taking these constant quantities outside of the integral, and writing down sinθ and r, in terms of these quantities . . .
And here, let’s also make another change of notation and, since this incremental displacement vector, dl, represents a displacement along y axis, this quantity can also be represented as dy in this, our new coordinate system. So, B becomes equal to µ0 i over 4π integral of, we will replace dl, incremental displacement vector magnitude with dy along that coordinate system, and, divided by r2, and if you take the square of r, we will have y2 plus r2, and then, this will be multiplied by sinθ.
And sinθ is, this is also equal to sinπ minus θ. Let me just write down over here. Sinθ is equal to sinπ minus θ. And, so if we express sinθ in its explicit form, that will be equal to R over square root of y2 plus R2. If we go back our diagram, now since we defined this axis as y, we said that idl is equal to i dy now, because dy is a small increment to this y coordinate of this specific point here. And, we’re going to integrate this throughout the whole length of this current-carrying wire.
Our variable is y, and it is going to go, therefore, from minus infinity to plus infinity. A moment ago, we have found the directions of these magnetic field vectors generated by these incremental current elements. And, since the field-generated by the upper half adds to the lower half, then we can write down this integral, also, as µ0 i over 4π, and take the integral from 0 to infinity, and multiply it 2. 2 times integral from 0 to infinity of R dy over y2 plus r2 to the power 3.5.
Now, the task becomes taking this integral. Earlier, we have seen exactly a similar type of integral when we were studying the electric field. And, we cannot make the usual u transformation for this type of integral, because, if we call the denominator in the parentheses as u, then du will be 2y dy. Since we don’t have any y term in the numerator, then we cannot apply that transformation. And, as you recall for this type of integral, we said that we use a trigonometric substitution.
And, we said that let y2 plus, or actually, let y is equal to RtanΦ, where Φ is our new variable. Then, dy becomes equal to R over cos2Φ d Φ. The derivative of tangent is secant square, and secant is 1 over cosine. Well, and, then if you try to represent this denominator in terms of new variable, y2 plus R2 to the power 3.5 is going to be equal to, for y2, we will have R2 tan2Φ, and then plus R2 to the power of 3.5.
Which will be equal to, R2 is common, so we can take it into R2 parentheses, and we’re going to have tan2Φ plus 1 to the power 3.5. We can take R2 outside of the power brackets. Simply, when we do that, we multiply the powers. Therefore, R2 is going to come out as R3, and inside of the power brackets, for tan2, we can use the sin2 over cos2 Φ as equal to tan2Φ and plus 1, again, to the power 3.5.
And if we have common denominator for the terms inside of the bracket, then we will have sin2Φ plus cos2 divided by cos2Φ to the power 3.5. And moving on, y2 plus R2 to the power 3.5 will be equal to R3 times sin2 plus cos2 is equal to 1. And in the denominator, we have cos2 Φ to the power 3.5. Again, if you take the quantity outside of this power bracket, we simply multiply the powers.
Of course, the 1 to the power 3.5 is going to give us just 1. So we will have R3 from this term, and if you take cos2 out, it will, it’s going to come out as cos3Φ. So, the denominator of the integral will be equal to this quantity in terms of this new variable Φ, and the numerator is going to be equal to R times this quantity. All right, then, the magnetic field will be equal to µ0 i over 4π, and we had this 2, let’s not forget that over there. 2 and, expressing the integrand in terms of the new variable, 4 . . .
Let’s see, R dy term, which we can take now, R is outside. For dy term, we have R over cos2 Φ d Φ divided by y2 plus R2 to the power of 3.5, and that is this term over here. We have R3 over cos3Φ. Now, since we changed the variable from y to Φ, then the boundaries of the integral will also change. We’re not going to calculate those boundaries because eventually, we’re going to go back to the original variable of y.
And here, if we do some cancellations, R times R will make R2, and we have R3 in the denominator; therefore, we’re going to end up only with R in the denominator. cos2 will cancel cos3, which will give us only 1 over cosΦ in the denominator. So, B is going to be equal to, here let’s also cancel this 2 with this 4, leaving us 2 in the denominator, and then take this R outside of the integral because it is constant, we will have µ0i over 2π, and times R.
Inside of the integral, we will have, this 1 over cosΦ will go as cosΦ to the numerator, so we’ll end up with cosΦ d Φ, integrated from Φ 1 to Φ 2. So, the integrant induces into this very simple form, and, we know that the integral of cosΦ is sinΦ; therefore, the magnetic field becomes equal to µ0 i, divided by 2πR times sinΦ, which will be evaluated at Φ 1 and Φ 2. We’d like to go back to our original variable of y. In order to do that, we will use the definition of our new variable.
And if you recall that the y was defined as, in terms of this new variable, as RtanΦ. So, the tanΦ will be equal to y over R. If we construct a right triangle which will satisfy this relationship, if this angle is Φ, then the opposite side should be y, and the adjacent side should be equal to capital R. So that the tanΦ will be equal to y over R. If these two sides are y and R, then the hypotenuse of this right triangle will be equal to square root of y2 plus R2.
Now, we can determine from this triangle that the sinΦ is equal to ratio of opposite side, which is y, to the hypotenuse, and that is square root of y2 plus R2. Okay. Then, the magnetic field will be equal to µ0 i over 2πR times sinΦ, which is y over square root of y2 plus R2, which will be evaluated at our original boundaries, and they were 0 and infinity. Well, we’re going to substitute first infinity for y, and then 0 for y, and we will look at the different to be able to get the final result.
If you substitute infinity to this equation for y, we will have infinity in the numerator, and then we’re going to end up with infinity in the denominator, and, from calculus, we know that infinity over infinity is undefined. So we have to make one more manipulation to our equation to substitute these boundaries. And the way that we will do that is, we will take the y2 term outside of the bracket. Doing that, we will have µ0 i over 2πR. In the numerator, we have y. If you take y2 out of square root, it will come out as y.
And in the inside of the square root, we will have 1 here, and since R2 doesn’t have any y2 multiplier, we have to divide that quantity by y2 to be able to have the same function. And we will evaluate this at 0 and infinity. Now, doing this, we will see that these y’s will cancel. If we substitute infinity for y here, a number divided by infinity will go to 0, so we end up only 1 in the square root, which will come out as 1, and 1 over 1 will give us just 1.
Therefore, the first boundary will give us 1 for that term, and if you substitute 0 for y, we will have a number divided by 0. Number divided by 0 will go to infinity. You can easily recall these things by visualizing that you’re dividing a number by a very, very small number, and obviously, the smaller the denominator, then the ratio will become larger and larger. So, if this term goes to infinity, then we will have infinity in the whole denominator. A number divided by infinity will go to 0.
So, the second boundary will eventually lead us to 0. Then, the magnitude of the magnetic field generated by this infinitely long, straight current-carrying wire, will be equal to µ0 i over 2πR. And the direction, if we just express this in vector form, at that point, will be into the plane as we have determined earlier by applying right-hand rule. Let’s go back to our diagram over here. As you can see, at this point, the magnetic field, the net magnetic field that this current-carrying wire generates as the current flows through along this wire, will be into the plane, pointing into the plane.
Now, if you consider a symmetric point, on the other side of the wire, somewhere over here, for example . . . Let’s call it p prime here . . . The associated position vectors relative to this point, from idl, is going to be pointing this way. From the lower idl, is going to pointing this way. And if we apply idl cross r, curling the right-hand fingers from idl towards r, and keeping the right-hand thumb in up position, we will see that the magnetic field from the upper part is going to be pointing out of plane.
And similarly, for the lower part, holding first right-hand fingers in the same direction of this idl, and then adjusting them and curling them towards the vector r over here, we will see that, again, the right-hand thumb will be pointing out of plane. So, from the lower part, also, the magnetic field will be pointing out of plane at this point. So here, magnetic field is into the plane, and whereas here, it is out of plane. Well, if you look at the points R distance away from this wire, they’re going to be, naturally, falling along a circle with the radius R, and calculate the magnetic field.
Let’s for example, consider a point over here, at this arbitrary location. Again, it is R distance, R distance away from the wire, and, for this point, the position vector will be drawn from idl to the point of interest pointing like this, and idl is up. And, as you recall from the cross product properties, that the resultant vector from a cross product is always perpendicular to those two vectors, which we’re taking cross product with, we can determine or easily see which direction that the resultant vector over here is going to be pointing.
So idl is up, r is in this direction, and if you take the cross product of idl and B, curling the right-hand fingers from idl towards this r, idl and r, I should have said. And, the resultant vector will be perpendicular to both of these two vectors. Well, in that case, first of all, we know that, to be able to perpendicular to this r, it will be tangent to this circle. And also, it has to be perpendicular to the direction of idl. So it’s going to be pointing, then, in a direction, something like this. All right. And same thing, we will end up from the lower part, that we’re generating over here.
If, then, I look at that case, from the top view . . . Let me just redraw that diagram over here. This is from the side view, that the current is moving up, and we have seen that R distance away, the magnetic field is pointing into the plane, and, by looking at the symmetrical point over here, we have found that the magnetic field is, at this end, coming out of plane. So, if we look at the top view, then we will see that the current is coming out of plane. That represents that the current is coming out of plane.
And, remember we’re looking at the top view from here, and our point of interest is over here. And at that point, the magnetic field is pointing into the plane from the side view, and from the top view, it will be pointing like this. The symmetric one, on the other side, the magnetic field is in opposite direction to that one, therefore it’s going to be pointing like this. Now, as we go along a circle with radius R, around this wire, we said that at an arbitrary location somewhere over here, the magnetic field is pointing like this. So now, we see the trend.
Since we know that the magnetic field is tangent to the field line passing through the point of interest, and as long as we’re this R distance away from the wire, we see that this current-carrying wire is going to be generating a magnetic field line, which is going to be circling like this in counter-clockwise direction. And the magnetic field at any point along this field line will be tangent to the field line. So, if we’re interested with the magnetic field of a point over here, or the magnetic field generated by the current-carrying wire at that point, will be tangent to the field line passing through that point.
Here, it is going to be tangent to that field line passing through that point, and so on, and so forth. So, by looking at this diagram, then, we can say that, one can easily determine the magnetic field generated by a current-carrying conductor at a specific point. Applying the right-hand rule in a different way to the case, and that right-hand rule, for a straight current-carrying wire, is such that . . . Let’s write this down again, step by step. First, hold your right-hand thumb in the direction of current flow and, as a second step, curl your right-hand fingers about the wire.
That is also the same axis with your right-hand thumb, or about the right-hand thumb. The direction of rotation of your right-hand fingers gives the direction of concentric magnetic field lines generated by the current-carrying wire. And, since now we know that the magnetic field is tangent to the field line passing through the point of interest, then the magnetic field at a specific point will be tangent to this circular field line generated by this current-carrying conductor.
So, the magnetic field is tangent to the field line passing through the point of interest. Therefore, if you go back to our diagram, then if the current is coming out of plane, in other words, when we look at the wire from the top view, by holding the right-hand thumb in the direction of flow of current, the field lines are going to be in the form of concentric circles. So, in a way, for a three-dimensional case, the field line is going to be circling, in counterclockwise direction, going into the plane here, and moving behind the plane, and coming out of over here.
And the magnetic field is tangent to the field line passing through the point of interest. And, so these field lines are, basically, in the form of concentric circles, rotating in counterclockwise direction for this case. So, from the top view, then the field lines are going to be something like this, in the form of concentric circles rotating in counterclockwise direction. And, if we’re interested in the magnetic field at this point, for example, right here, and the field line passing through that point is this one, then the B field will be tangent to the field line at that point.
So it’s going to be, then, something like this. And the magnitude of the field line, for a wire like this, is going to be equal to this quantity. Actually, let me just write down the magnitude over here. B will be equal to µ0i over 2π. Whatever that distance to the point of interest is. Well, this is for the case, if the current is coming out of plane. If the current is going into the plane, applying the right-hand rule, we will hold the right-hand thumb in the direction of flow of current now.
It is pointing into the plane, and as we hold the right-hand thumb pointing into the plane, and curling the right-hand fingers about the thumb, that will give us the sense of rotation of the magnetic field lines, concentric magnetic field lines, and in that case, they’re going to be in clockwise direction. So we can summarize those cases over here. And, if the current is coming out of plane, the B field lines are in the form of concentric circles.
And keeping the, holding the right-hand thumb out of plane direction, and curling the right-hand fingers, they will generate counterclockwise magnetic field lines. The magnetic field will be tangent at the point of interest to the field line passing through that point. And, if the current is going into the plane, again, the field lines are going to be in the form of concentric circles. But in this case, they’re going to be rotating in clockwise direction, once we apply the right-hand rule.
If you hold your thumb into the plane, and curl your fingers, they will curl in clockwise direction. And again, the magnetic field will be tangent to the field line passing through the point of interest. And the magnitude of this field line is equal to µ0 i over 2πR distance. R is the distance between the wire and, let’s use capital R over here, distance between the wire and the point of interest for this specific point.