Figure 30-45 shows a cross section of a long cylindrical...

Figure $30-45$ shows a cross section of a long cylindrical conductor of radius a containing a long cylindrical hole of radius $b .$ The axes of the cylinder and hole are parallel and are a distance $d$ apart; a current $i$ is uniformly distributed over the tinted area. (a) Use superposition to show that the magnitude of the magnetic field at the center of the hole is $$ (b) Discuss the two special cases $b=0$ and $d=0$. (Hint: Regard the cylindrical hole as resulting from the superposition of a complete cylinder (no hole) carrying a current in one direction and a cylinder of radius $b$ carrying a current in the opposite direction, both cylinders having the same current density.) B=\frac{\mu_{0}|i| d}{2 \pi\left(a^{2}-b^{2}\right)} $$

Figure $30-45$ shows a cross section of a long cylindrical conductor of radius a containing a long cylindrical hole of radius $b .$ The axes of the cylinder and hole are parallel and are a distance $d$ apart; a current $i$ is uniformly distributed over the tinted area.
(a) Use superposition to show that the magnitude of the magnetic field at the center of the hole is
$$
(b) Discuss the two special cases $b=0$ and $d=0$. (Hint: Regard the cylindrical hole as resulting from the superposition of a complete cylinder (no hole) carrying a current in one direction and a cylinder of radius $b$ carrying a current in the opposite direction, both cylinders having the same current density.)
B=\frac{\mu_{0}|i| d}{2 \pi\left(a^{2}-b^{2}\right)}
$$

Key Concepts

Analysis of Special Cases

Exploring special cases, such as when the hole radius or the distance between the cylinder and the hole center is zero, serves as an important consistency check in electromagnetic problems. By verifying that the solution simplifies to known results in these limiting cases, one gains assurance that the general approach is correct and that the derived expression properly accounts for all physical parameters.

Magnetic Field in Cylindrical Geometries

The determination of the magnetic field in cylindrical geometries often involves applying Ampère’s law or the Biot–Savart law, exploiting the symmetry of the system. In a cylindrical conductor, the radial dependence of the magnetic field can be calculated analytically, and the use of symmetry simplifies the problem considerably, especially when dealing with uniform currents.

Treatment of Holes and Negative Current Density

When a cylindrical hole is present in a conductor, it can be mathematically modeled by superimposing a region of negative current density over an otherwise full conductor. This method effectively subtracts the contribution of the missing current-carrying material, allowing for the calculation of the magnetic field in regions affected by the hole.

Superposition Principle

The superposition principle in electromagnetism states that the total magnetic field from several sources is the vector sum of the fields produced by each source individually. In problems involving complex geometries, such as a conductor with a hole, this principle allows one to decompose the system into simpler parts—like a complete conductor and an oppositely directed fictitious current representing the hole—and then combine the results to find the net field.

Uniform Current Distribution

Uniform current distribution refers to the condition where the current density is constant over the cross-sectional area of a conductor. This concept simplifies the computation of magnetic fields and other electromagnetic quantities because it ensures that the current is spread evenly, allowing one to apply symmetry arguments and standard integration techniques in cylindrical geometries.

Transcript

00:01 In this problem, we have a large cylinder with the current i distributed not over the whole cylinder, but over only what i've crosshatched.

00:10 There is a whole radius b where there's no current flowing.

00:15 And the axes of the two cylinders are parallel, and the senders are separated by distance d.

00:22 So let's get the current density.

00:25 J is equal to, this is the magnitude of i over pi.

00:32 A squared minus pi b squared because it's not over the whole area got that hole so that's our current density our magnitude our current density now how are we going to analyze this problem well we're going to think of current the larger cylinder as having current flowing through all of it now with that current density because we got to have everywhere other than the hole having that it's got to be the same problem but then we add to it we add to it we add to it a solid cylinder with current flowing the opposite direction.

01:07 So in the large cylinder, it's out towards you.

01:10 In the smallest cylinder, it's into the screen.

01:12 When you put these two together, they cancel each other out in that region.

01:17 So we do get back to the original picture.

01:21 You do get back.

01:22 So now let's look at the ampios law for a large cylinder, get the magnetic field, sense of magnetic field.

01:31 Now, here's my mp.

01:34 Bearing loop, i'm going to go around it counterclockwise.

01:40 The b .ds, that actually gives you the s component.

01:45 What is the s component? that tells you, are you in the same direction as ds or the opposite direction of ds? every point, you have that.

01:56 And one point is going to be no different than the other.

01:59 It's a uniform distribution of current.

02:02 So i can factor that out.

02:03 B -s -i -s -equal -d -s, is equal to mu -knot.

02:10 Now, the current enclosed, fingers wrapping around the curve in the sense of the arrow, thumb is out.

02:16 That's positive current.

02:17 Notice i've made the largest cylinder and have the current coming out towards you.

02:21 So, mu -not, j, pi, r square.

02:26 That's what we get for the enclosed current.

02:29 So this gives me that bs 2 pi -r.

02:36 Is equal to mu not, and then we have i pi r squared over pi a squared minus b squared.

02:49 So, pi goes away here.

02:52 This r gets rid of the square.

02:55 So we get from this, b s is equal to mu not, i, r, 2 pi, a squared, minus b squared.

03:06 This is greater than zero.

03:09 That tells me that at the center of the hole, and let me put an l on this to indicate this is for the large cylinder.

03:21 That bl is upward the center of hole because you're in the same direction.

03:37 You're the same direction as ds, the center of the hole.

03:40 Think about this curve being now through the center of the hole.

03:43 Ds would be straight up.

03:45 So bs is positive.

03:46 That means b itself is upward.

03:50 Now, for the small cylinder, nothing really special happens here because we've already done it for the large one.

04:04 This is going to be mu not, i, r prime...

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