Using Ampere's circuital law, calculate the magnetic induction due a cylindrical current carryin

step 1 In this problem, we're going to talk about compares law. So what we need to remember is that if

Using Ampere's circuital law, calculate the magnetic...

Key Concepts

Current Distribution in Cylindrical Conductors

In problems involving cylindrical conductors, it is often assumed that the current is uniformly distributed over the cross-sectional area of the conductor. This uniform current distribution enables the use of symmetry to simplify calculations, allowing for the determination of the enclosed current by the chosen Amperian loop and ultimately leading to an analytical expression for the magnetic field at various points, including on the surface of the conductor.

Magnetic Induction

Magnetic induction, commonly represented by the magnetic field vector, describes the influence of electric currents and magnetic materials on the space around them. It is a vector quantity that encapsulates both the strength and the direction of the magnetic influence, and is a core concept for understanding how steady currents generate magnetic fields, as captured by Ampere's Circuital Law.

Symmetry in Electromagnetic Problems

Exploiting symmetry is essential in electromagnetic calculations, as seen in the analysis of the magnetic field around an infinitely long cylindrical conductor. High symmetry (such as cylindrical symmetry) simplifies the integration process by reducing complex field distributions to manageable forms where the magnetic field has a constant magnitude along an appropriately chosen path, making mathematical evaluation more straightforward.

Ampere's Circuital Law

Ampere's Circuital Law is a fundamental principle in magnetostatics which relates the integrated magnetic field around a closed loop to the electric current passing through the loop. This law simplifies the calculation of magnetic fields in cases with high symmetry by allowing one to equate the circulation of the magnetic field to the current multiplied by the permeability of free space, giving a direct link between current and magnetic field.

Transcript

00:01 In this problem, we're going to talk about compares law.

00:03 So what we need to remember is that if we have a distribution of current, as i'm showing here on the screen, such that the current density per cross -sectional area actually is this vector j, then if we draw a circuit around any place, we obtain that the integral over the circuit of b so b times d l d l is the element of length is equal to mu zero the magnetic constant times the integral over the area enclosed by the circuit so this area here of j d s where the s is the element of of area.

01:06 This is in paris law.

01:09 And in our problem, we have a cylindrical conductor that has a radius r.

01:19 So like this, the radius is r.

01:26 But j, the current density is equal to b times r.

01:34 So we don't have a uniform current density.

01:39 And our goal with this and this is to find what is the value of the magnetic field b, a, when r is smaller than capital r, and b when r is greater than capital r.

02:04 Okay, so what we need to do first is to notice that j is pointing in this direction, it's flowing along the our cylinder and in question a we're going to use that the integral of b d l and i'm going to choose i'm going to choose this circuit right here okay that is a circle around that wraps around the cylinder with a radius but actually uh in our case for question a they should be in the circuit.

02:58 Okay, for question b, we're going to make it outside.

03:01 So this radius here is r.

03:05 So bdl over integrated over this circuit is equal to the integral of jds.

03:15 Okay.

03:16 I noticed that i'm, i just got rid of the vector notation because b must be perpendicular to dl since dl is in the same direction as the current, and b, by the right -hand rule, must be pointing perpendicularly to the current at all times here.

03:42 Actually, b will make a circle around our cylinder.

03:52 Okay, furthermore, we have that j and d .s are also perpendicular because d .s is, actually, i'm sorry, they're not perpendicular.

04:07 They are parallel, and as a matter of fact, dl is also parallel to b...

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