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Chapter 5: Q5P (page 222)

A current Iflows down a wire of radius a.

(a) If it is uniformly distributed over the surface, what is the surface current density K?

(b) If it is distributed in such a way that the volume current density is inversely

proportional to the distance from the axis, what is J(s)?

Short Answer

Expert verified

(a) The surface current density corresponding to a current Iflowing down a wire of radius aisI2πa.

(b) The volume current density corresponding to a current Iflowing down a wire of radius ais I2πas.

Step by step solution

01

Given data

There is a wire of radius a carrying a current I.

02

Surface and volume current density

The surface current density if the current is uniformly distributed over a surface is

K=IL.....(1)

Here, L is the cross sectional length of the surface.

The total current in terms of volume current density is

I=∫Jda.....(2)

Here, the integration is over the cross sectional area.

03

Surface current density in wire of radius   

From equation (1), the surface current density on the surface of the wire of radius a is

K=I2πa

Thus, the uniform surface current density is I2πa.

04

Volume current density in wire of radius  

Let the volume current density be

J=ks.....3

Here, k is a constant and is the radial component of polar coordinate.

From equation (2),

I=k∫0a1s2πsds=k2πak=I2πa

Substitute this in equation (3) to get

J=I2πas

Thus, the volume current density is I2πas.

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Most popular questions from this chapter

A thin glass rod of radius Rand length Lcarries a uniform surface charge σ. It is set spinning about its axis, at an angular velocityω. Find the magnetic field at a distances s冈Rfrom the axis, in the xyplane (Fig. 5.66). [Hint: treat it as a stack of magnetic dipoles.]

(a) Prove that the average magnetic field, over a sphere of radius R,due to steadycurrents inside the sphere, is

B→ave=μ04π2m→R3

wherem→is the total dipole moment of the sphere. Contrast the electrostatic

result, Eq. 3.105. [This is tough, so I'll give you a start:

B→ave=143πR3∫B→dτ

WriteB→as∇→×A→ ,and apply Prob. 1.61(b). Now put in Eq. 5.65, and do the

surface integral first, showing that

∫1rda→=43πr'

(b) Show that the average magnetic field due to steady currents outsidethe sphere

is the same as the field they produce at the center.

Question: (a) Find the density ρof mobile charges in a piece of copper, assuming each atom contributes one free electron. [Look up the necessary physical constants.]

(b) Calculate the average electron velocity in a copper wire 1 mm in diameter, carrying a current of 1 A. [Note:This is literally a snail'space. How, then, can you carry on a long distance telephone conversation?]

(c) What is the force of attraction between two such wires, 1 em apart?

(d) If you could somehow remove the stationary positive charges, what would the electrical repulsion force be? How many times greater than the magnetic force is it?

A circularly symmetrical magnetic field ( B depends only on the distance from the axis), pointing perpendicular to the page, occupies the shaded region in Fig. 5.58. If the total flux (∫B.da) is zero, show that a charged particle that starts out at the center will emerge from the field region on a radial path (provided it escapes at all). On the reverse trajectory, a particle fired at the center from outside will hit its target (if it has sufficient energy), though it may follow a weird route getting there. [Hint: Calculate the total angular momentum acquired by the particle, using the Lorentz force law.]

A steady current Iflows down a long cylindrical wire of radius a(Fig. 5.40). Find the magnetic field, both inside and outside the wire, if

  1. The current is uniformly distributed over the outside surface of the wire.
  2. The current is distributed in such a way that Jis proportional to s,the distance from the axis.
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