The law of Biot and Savart in Eq. ( 28.7 ) generalizes to...

The law of Biot and Savart in Eq. ( 28.7 ) generalizes to the case of surface currents as $$ \overrightarrow{\boldsymbol{B}}=\frac{\mu_{0}}{4 \pi} \int \frac{\sigma \overrightarrow{\boldsymbol{v}} \times \hat{\boldsymbol{r}}}{r^{2}} d a $$ where $\sigma$ is the local charge density, $\overrightarrow{\boldsymbol{v}}$ is the local velocity, and $d a$ is a differential area element. Re-visit Challenge Problem 28.76 and use the above equation as an alternative means to derive the magnetic field at the center of the cylinder. Use the following steps: (a) Write the charge density $\sigma$. (b) The origin is at the center of the cylinder. What is the vector $\vec{v}$ that points from the element with coordinates $(x, y, z)=(x, R \cos \phi, R \sin \phi)$ to the origin? (c) What is the velocity $\overrightarrow{\boldsymbol{v}}$ of the element? (d) What is the vector product $\overrightarrow{\boldsymbol{v}} \times \hat{\boldsymbol{r}} ?$ (e) An area element on the cylinder may be written as $d a=R d x d \phi .$ Use this and the previously established information to write the generalized law of Biot and Savart as a double integral. Evaluate the integral to determine the magnetic field $\vec{B}$ at the center of the cylinder. (f) Is your result consistent with your result in Challenge Problem $28.76 ?$

The law of Biot and Savart in Eq. ( 28.7 ) generalizes to the case of surface currents as
$$
\overrightarrow{\boldsymbol{B}}=\frac{\mu_{0}}{4 \pi} \int \frac{\sigma \overrightarrow{\boldsymbol{v}} \times \hat{\boldsymbol{r}}}{r^{2}} d a
$$
where $\sigma$ is the local charge density, $\overrightarrow{\boldsymbol{v}}$ is the local velocity, and $d a$ is a differential area element. Re-visit Challenge Problem 28.76 and use the above equation as an alternative means to derive the magnetic field at the center of the cylinder. Use the following steps: (a) Write the charge density $\sigma$. (b) The origin is at the center of the cylinder. What is the vector $\vec{v}$ that points from the element with coordinates $(x, y, z)=(x, R \cos \phi, R \sin \phi)$ to the origin? (c) What is the velocity $\overrightarrow{\boldsymbol{v}}$
of the element? (d) What is the vector product $\overrightarrow{\boldsymbol{v}} \times \hat{\boldsymbol{r}} ?$ (e) An area element on the cylinder may be written as $d a=R d x d \phi .$ Use this and the previously established information to write the generalized law of Biot and Savart as a double integral. Evaluate the integral to determine the magnetic field $\vec{B}$ at the center of the cylinder. (f) Is your result consistent with your result in Challenge Problem $28.76 ?$

Key Concepts

Vector Cross Product and Direction of Magnetic Field

The vector cross product is a mathematical operation that yields a vector perpendicular to the plane formed by two input vectors. In the context of the Biot-Savart law, the cross product between the velocity (or current) vector and the unit vector from the current element to the observation point determines both the magnitude and the direction of the associated magnetic field contribution. This concept is pivotal for ensuring that the magnetic field’s direction obeys the right-hand rule and correctly reflects the physical behavior of magnetic fields generated by moving charges.

Generalized Biot-Savart Law

The Biot-Savart law is a fundamental principle used to calculate the magnetic field generated by a current distribution. Its generalized form extends to current distributions that exist on surfaces (or volumes) by integrating over the appropriate differential elements. The law in this context involves summing contributions from infinitesimal current elements, with the magnetic field from each element determined using the cross product of the current vector and the unit position vector, scaled by the inverse square of the distance between the element and the observation point.

Surface Charge and Current Density

In the context of magnetic fields generated by moving charges on a surface, the surface charge density represents the distribution of electric charge per unit area. When these charges are in motion, they create a surface current density, which is obtained by multiplying the surface charge density by the local velocity of the charges. This concept is crucial because it provides a link between the static distribution of charge and the dynamic magnetic field generated by their motion.

Differential Area Element Integration

Evaluating magnetic fields from extended current distributions requires breaking the charged surface into infinitesimal differential area elements. Each of these elements contributes to the overall magnetic field, and integration over the entire surface allows for the summation of these individual contributions. Understanding the appropriate expression for a differential area element, especially in non?Cartesian geometries like cylindrical surfaces, is essential for setting up the integral in the Biot-Savart law.

Transcript

00:01 Here we're going to calculate the magnetic field right at the center of a rotating cylinder that holds a surface charge.

00:10 So this is a similar construction to a solenoid, and we will check our answer using a solenoid magnetic field when we get done.

00:21 But there are some shortcuts that i'm going to take, but we'll start with the full version of the biotse of art law.

00:30 And if you remember, there was a q in the original formulation of this.

00:38 It was mu knot over 4 pi, was the magnetic field.

00:42 And then q, v crossed r hat over r squared.

00:49 That was, of course, for a point charge.

00:52 And here we're making the replacement of q with a surface charge density integrated over the surface area of, in this case, a cylinder.

01:11 We will be assuming that that surface charge density is constant, and that will make life a little bit easier.

01:20 The other thing that i have done to make life a little bit simpler is instead of using a unit vector on r, i use the fact that r vector is equal to r -hat, the unit vector, over the magnitude, sorry, that's just the opposite of that.

01:41 Let's write that the right way, that the unit vector can be replaced by the full vector r divided by its magnitude.

01:51 And so i have substituted r hat with that particular form.

01:58 The last thing to realize before we get into the nitty -gritty is that the vector r is actually a difference in two position vectors.

02:09 It is the observation point, which is held fixed, minus the source point, which will vary as you integrate over that surface.

02:24 You are changing what you consider to be the source point.

02:29 So to demonstrate this, i have a little strip dx, and i'm going to be making some simplifying assumptions because of the symmetry in that little strip that i've shown.

02:42 But your source point is inside the cylinder, and so the r vector actually points from that source point to the origin.

02:56 So that is the sense of what r is.

03:01 Now we can write down the coordinates of both our source point and our observation point fairly easily.

03:11 I'll give myself some room here.

03:15 The observation point, of course, is the origin.

03:19 So it's 0 -0 .0.

03:22 Let me just write that as a threesome rather than with the ijk or x, y, k, or x, y, z notation.

03:31 The source point is you're somewhere out along the x axis, along the circle, and i've shown that circle in blue, so i imagine that running a circle all the way around.

03:48 We have r cosine of fee and r sign of fee, where fee, or phi, if you want to call it, that is the asmusal angle starting.

04:03 I should have a coordinate system there.

04:05 X is axial.

04:08 Y i'm considering to be horizontal in the plane into the plane of the paper, and z to be up, just to pick a coordinate system.

04:22 Okay, so we can write down fairly quickly what r is.

04:26 It unfortunately has three negative components to it.

04:32 But what you notice is that there is a high degree of symmetry.

04:40 So what you would expect is that there should be only one component of the magnetic field.

04:52 So the phi symmetry indicates that there should just be an axial magnetic field.

05:01 This would not be true if, for example, half the cylinder were cut away or a portion of the cylinder or somehow cut away.

05:13 But because it is the same all the way around, we only expect a bx component.

05:24 And so we are going to be looking for the cross product that has just the x component that we worry about in v -crossed r.

05:36 Okay, so here we're going to go through the details.

05:39 But a reminder that we're only finding bx and we expect by equal to bz equal to zero.

05:56 And if you were confident about that, you could include those components and do the integrals and you would quickly see that it was a lot of effort, all for literally not.

06:12 So anyway, we have r.

06:13 We can just take the pythagorean theum, the sum of the squares of the components of our vector, any negative signs will, of course, go away with the squared, and of course the sine squared plus cosine squared is going to equal one.

06:44 So that will simplify down, square root of x squared plus i squared.

06:53 The other thing we need is to figure out the cross product, and then we will be ready to set up the integral.

07:04 And the cross product, we do need an expression for v in order to do the cross product between our two vectors.

07:15 And i will draw v somewhere in that circle.

07:21 It's probably a good thing to draw it where the y, the y, is a positive quantity, so maybe over on the other side of the circle.

07:36 V is pointing up initially at phi equals to zero.

07:47 So we'll kind of do what we do with the x -axis, but call the y -axis, phi -equal zero.

07:55 And so you have the biggest y component, which would be of magnitude v times cosine of f, and v, by the way, we can use a little bit of rotational motion and understand that that is a tangential velocity.

08:16 So it is going to have magnitude equal to the rotational velocity times the radius.

08:25 It will have no x component, but we'll have a y component initially positive.

08:33 And 90 degrees later, it will have a.

08:37 Z component.

08:43 Actually, i just have that reverse, don't i? okay, let's go back and look at the coordinate system.

08:52 Make myself slow down a little bit.

08:54 Yeah, so initially it has all positive z component at phi equals to zero, and 90 degrees later, it will have a negative y component.

09:07 So yes, let's get that all cleared up.

09:21 Okay, and we should check.

09:23 That should be perpendicular to the r vector, which it is.

09:33 So if you take the dot product between v and r, you should get zero.

09:41 And so the cosine, sign terms will actually provide a way to turn that into zero...

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