A conductor wire is in the form of a closed circle. In this series, we will learn to obtain the magnetic field of a point on the axis of the circle. The magnetic field is calculated using the Biot-Savart law.
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00:00Hi colleagues, meet again with our channel, DigoScience.
00:05A channel that presents educational videos in animation form.
00:11Let's go into the material.
00:14The amount of magnetic induction at a point around a conductor carrying an electric current can be calculated using the Biot-Savart law.
00:21dB is equal to µ0 over 4i, multiplied by idl sine theta over r squared.
00:30We will apply this equation to a circular conductor.
00:35Before going any further, place this circle into a coordinate system.
00:41Why? Because magnetic induction is a vector quantity.
00:46It is easier to analyze a vector quantity if the vector is in a coordinate system.
00:54Let's assume the circle is in the x-y plane, such that the center of the circle is at the coordinate center.
01:01So, we want to calculate the magnitude of the magnetic induction from a point located on the z-axis.
01:08The point that is on the axis of the circle.
01:13In the conductor, electric current flows counterclockwise.
01:21We already know Biot-Savart's law very well.
01:25Let's assume that the conductor is composed of several small segments.
01:29Let's start from this small segment. Small segment that is on the y-axis.
01:36The length of this small segment is dl.
01:41Because the direction of the dl vector is in the same direction as the direction of the electric current,
01:46the direction of the dl vector is to the left of the x-axis, or to the left.
01:53Connect small segments with points on the coordinate axis.
01:57The length of this line is r.
02:01The unit vector r-hat is in line with that line.
02:05The direction of the r-hat vector is towards the test point.
02:11We don't know the radius value of the circle yet. Think of it as a big RnL.
02:19We will predict the magnetic field vector.
02:22This is complicated enough if you can't imagine vector product.
02:30As with other vectors, you can shift the dl vector and r-hat vector to the test point.
02:38The dB vector will be perpendicular to the dl vector and the r-hat vector.
02:46Here's the dB vector.
02:50Well, this test point gets the magnetic field from each small segment of the circle.
02:57Several vectors acting at the same point have a resultant vector.
03:01This is a concept that has been understood for a long time, right?
03:07Let's look at this image in the y-z plane.
03:11Small segments of current are out of the image plane.
03:14We can write it as a point in a circle.
03:20To analyze dB, let's assume this angle is angle alpha,
03:26Angles that are opposite each other have the same angle value.
03:31Thus, this angle is alpha.
03:36Remember vector dB is perpendicular to vector dl and vector r-hat.
03:41This angle is 90 degrees.
03:44If so, this angle is 90 degrees minus alpha.
03:51The z-axis and y-axis also form a right angle.
03:55That is, this angle is alpha.
04:00Knowing the value of alpha we can decompose the vector dB into a component vector on the z-axis, dB sine alpha,
04:07and the component vector on the y-axis, dB cosine alpha.
04:11At this stage, the process is still understandable.
04:16A circle is a symmetrical object.
04:19When the current reaches the negative y-axis, the direction of the dl vector is into the plane.
04:27This small segment produces a dB vector like this.
04:33This dB vector can also be decomposed into component vectors.
04:39Now notice, the component vector dB cosine alpha has the same magnitude, but opposite direction.
04:46These two vectors will cancel each other out.
04:49What remains are two vectors dB sine alpha.
04:56Look at the animation as follows.
04:59The magnetic field at the test point only has a direction to the positive z-axis.
05:06Now we can write Biot-Savart's law as follows.
05:12The angle between the dl vector and the r-hat vector is 90 degrees.
05:17Then theta is 90 degrees.
05:23This factor is zero, because the components of this vector cancel each other out.
05:30What remains is this factor.
05:35Now, pay attention to a small segment.
05:38The small segment is as many as the circumference of the circle.
05:42The circular conductor is formed by small segments from zero to the circumference of the circle, 2Ï€R.
05:52The magnetic induction at the test point is the resultant of each dB produced by the small segment.
06:00The value of B at the test point is the integral of dl from zero to 2Ï€R.
06:08We can exclude all fixed values, or values that do not change from integral notation.
06:16The integral of 1 over dl is nothing other than L.
06:22Enter the integration value.
06:27B is equal to μ0 I big R sin α over 2R squared.
06:31This is the magnitude of the magnetic induction at the test point.
06:36The direction is upward, or the positive z-axis.
06:41We will explore this formula in the next series.
06:44Hope it is useful.
06:46And, don't forget to watch the next video.