Ssm A long, cylindrical conductor is solid throughout and...

ssm A long, cylindrical conductor is solid throughout and has a radius $R$ . Electric charges flow parallel to the axis of the cylinder and pass uniformly through the entire cross section. The arrangement is, in effect, a solid tube of current $I_{0} .$ The current per unit cross-sectional area (i.e., the current density ) is $I_{0} /\left(\pi R^{2}\right) .$ Use Ampere's law to show that the magnetic field inside the conductor at a distance $r$ from the axis is $\mu_{0} I_{0} r /\left(2 \pi R^{2}\right)$ . Hint: For a closed path, use $a$ circle of radius $r$ perpendicular to and centered on the axis. Note that the current through any surface is the area of the surface times the current density.)

Key Concepts

Symmetry in Electromagnetic Problems

Utilizing symmetry in electromagnetic problems simplifies the process of selecting an appropriate integration path, such as a circular loop in a cylindrical conductor. The symmetry ensures that the magnetic field is constant in magnitude and direction along the chosen path, thereby reducing the line integral to a simple product of the magnetic field magnitude and the path’s circumference.

Ampere's Law

Ampere's Law states that the line integral of the magnetic field around a closed path is proportional to the total electric current enclosed by that path. This fundamental principle in electromagnetism allows for the calculation of magnetic fields in situations with high symmetry, linking the circulation of the field directly to the enclosed current.

Current Density

Current density is defined as the electric current per unit area, representing how uniformly the current is distributed across a conductor's cross-section. In cases of uniform current distribution, the current density provides a straightforward method to compute the enclosed current for a given area within the conductor.

Enclosed Current Calculation

The concept of enclosed current pertains to the portion of the total current that lies within the area bounded by the chosen integration path. In uniformly charged or current-carrying conductors, this is determined by multiplying the current density by the area enclosed by the path, which is critical for applying Ampere's Law effectively.

Transcript

00:01 In this problem, we're asked to use amper's law to figure out how large the magnetic field is going to be within a wire at some particular distance from the center of the wire up to the surface.

00:13 If we have a uniform current distribution across the cross section of that wire.

00:21 So we know that we have some wire, and i'm going to draw it rather thick for illustrative purposes.

00:29 Of course, you know, when we just look at wires, they usually look pretty thin, but we'll draw this cross -sectional area of this wire here.

00:40 So overall, a current i not is going to be going throughout this entire wire, but that's distributed all completely throughout this wire.

00:53 So if i wanted to know how much current is going through, let's say, not the entire wire, right around all here, but say instead i wanted to know how much current was only going through this much, right, of wire, then i would need to know something about the current density, which is what we do know that.

01:19 So say we have this little bit, this little circle of little radius r here, right? imagine we wanted to know how much of the entire current ionod is going to be traveling through that bit.

01:31 Well, i would need to know this cross -sectional area of my little green circle here, and that's going to be pi r squared, of course.

01:39 And then i would need to know the current density.

01:42 So how much current will be encapsulated by each of these cross -sectional areas when i draw them.

01:52 So i've given that the current per cross -sectional area is i -0 divided by.

02:00 The entire cross -sectional area of the entire wire.

02:08 So i take this, multiply it by the cross -section i care about, and i get how much current is going through just that little square, right? so you'll see that if i did want to use the circle of all of this, well, what would i plug in for my radius? big r, well, if i plug that in there, i get that the current through all of the wire, right? is i0 over 2 pi big r squared, multiplied by pi, big r squared.

02:35 Oh, sorry.

02:36 The 2 pi.

02:37 I wrote by accident.

02:38 Let me erase that.

02:40 That 2 should not be there.

02:42 That's an accident there...

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