Question: The preceding problem was an artificial model for the charging capacitor, designed to avoid complications associated with the current spreading out over the surface of the plates. For a more realistic model, imagine thin wires that connect to the centers of the plates (Fig. 7.46a). Again, the current I is constant, the radius of the capacitor is a, and the separation of the plates is w << a. Assume that the current flows out over the plates in such a way that the surface charge is uniform, at any given time, and is zero at t = 0.

(a) Find the electric field between the plates, as a function of t.

(b) Find the displacement current through a circle of radius in the plane mid-way between the plates. Using this circle as your "Amperian loop," and the flat surface that spans it, find the magnetic field at a distance s from the axis.

Figure 7.46

(c) Repeat part (b), but this time uses the cylindrical surface in Fig. 7.46(b), which is open at the right end and extends to the left through the plate and terminates outside the capacitor. Notice that the displacement current through this surface is zero, and there are two contributions to Ienc.

Short Answer

Expert verified

Answer

(a) The value of the electric field between the plates, as a function of isE=ItSε0.

(b)

The value of displacement currentId=EdS .

The value of magnetic field at a distance s from the axis is B=μ0I2sAϕ^.

(c) The value of magnetic field in the same direction as the case b isB=μ0I2πsa2 .

Step by step solution

01

Write the given data from the question.

Consider that the current flows out over the plates in such a way that the surface charge is uniform, at any given time, and is zero at t = 0.

02

Determine the formula of electric field between the plates, as a function of t.

Write the formula of electric field between the plates, as a function of.

E=σtε0 …… (1)

Here, σt is charge on the plate and ε0is relative permittivity.

Write the formula of magnetic field at a distance s from the axis.

B=μ0Id2πs …… (2)

Here, μ0is permeability, Idis displacement current and s is the surface of the plates.

Write the formula of magnetic field in the same direction as the case b.

B=μ0I2πs1-IsI …… (3)

Here, μ0is permeability, I is current, IS is current passing through curve.

03

(a) Determine the electric field between the plates.

Determine the charge on the plate is:

Qt=σtS=It

Since the charge density grows linearly with time and is homogeneous throughout the plate. The electric field is thus:

Determine the electric field between the plates, as a function of t.

Substitute QtSforσtinto equation (1).

E=Qtε0S=ItSε0

Here, S is the surface of the plates.

04

(b) Determine the displacement current and magnetic field at a distance s  from the axis.

The magnetic field will curve around the current if it flows down the z-axis, using Ampere's equation with the displacement current term.

Determine the displacement current.

Id=EdS

Determine the magnetic field at a distance from the axis.

Substitute EDS for Id into equation (2).

B=μ0ε02πsddtSEdS=μ0s2π2πsIπa2=μ0I2sAϕ^

Whereas, the integration loop was positioned between the capacitor plates as a circle whose radius was coaxial with the plates.

05

(c) Determine the magnetic field in the same direction as the case b.

Using a new bounding surface but the same Amperian loop once more, the magnetic field is:

2πsB=μ0Ienc

There will be two contributions from "regular" current but there is contribution from the displacement current. Given that the surface's normal points inwards, one is on surface S1 (positive), and the other is at curve C1 (negative).

2πsB=μ0Ienc2πsB=μ0I-μ0IsB=μ0I2πs1-IsI

The charge density must only be a function of time for the current I(s) to travel through curve. Let's say that all we search for is the charge density on the circle enclosed by curve:

σt=Qs,ts2π=1s2πI-Ist

For to only be a function of time the term I-Is needs to be proportional to;

I-Is=Cs2

Substitute 0 for l(s) into above equation.

C=Is2

Now substitute 1s2for C into above equation.

Is=I-Isa2

Determine the magnetic field in the same direction as the case b.

B=μ0I2πs1-1+sa2=μ0I2πsa2

Therefore, the value of magnetic field in the same direction as the case b is .

B=μ0I2πsa2

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Most popular questions from this chapter

Sea water at frequency v=4×108Hzhas permittivitylocalid="1657532076763" =810, permeabilityμ=μ0, and resistivityρ=0.23Ω.m. What is the ratio of conduction current to displacement current? [Hint: Consider a parallel-plate capacitor immersed in sea water and driven by a voltageV0cos(2πvt) .]

The magnetic field outside a long straight wire carrying a steady current I is

B=μ02πIsϕ^

The electric field inside the wire is uniform:

E=Iρπa2z^,

Where ρis the resistivity and a is the radius (see Exs. 7.1 and 7 .3). Question: What is the electric field outside the wire? 29 The answer depends on how you complete the circuit. Suppose the current returns along a perfectly conducting grounded coaxial cylinder of radius b (Fig. 7.52). In the region a < s < b, the potential V (s, z) satisfies Laplace's equation, with the boundary conditions

(i) V(a,z)=Iρzπa2 ; (ii) V(b,z)=0

Figure 7.52

This does not suffice to determine the answer-we still need to specify boundary conditions at the two ends (though for a long wire it shouldn't matter much). In the literature, it is customary to sweep this ambiguity under the rug by simply stipulating that V (s,z) is proportional to V (s,z) = zf (s) . On this assumption:

(a) Determine (s).

(b) E (s,z).

(c) Calculate the surface charge density σ(z)on the wire.

[Answer: V=(-Izρ/πa2) This is a peculiar result, since Es and σ(z)are not independent of localid="1658816847863" zas one would certainly expect for a truly infinite wire.]

A familiar demonstration of superconductivity (Prob. 7.44) is the levitation of a magnet over a piece of superconducting material. This phenomenon can be analyzed using the method of images. Treat the magnet as a perfect dipole , m a height z above the origin (and constrained to point in the z direction), and pretend that the superconductor occupies the entire half-space below the xy plane. Because of the Meissner effect, B = 0 for Z0, and since B is divergenceless, the normal ( z) component is continuous, so Bz=0just above the surface. This boundary condition is met by the image configuration in which an identical dipole is placed at - z , as a stand-in for the superconductor; the two arrangements therefore produce the same magnetic field in the region z>0.

(a) Which way should the image dipole point (+ z or -z)?

(b) Find the force on the magnet due to the induced currents in the superconductor (which is to say, the force due to the image dipole). Set it equal to Mg (where M is the mass of the magnet) to determine the height h at which the magnet will "float." [Hint: Refer to Prob. 6.3.]

(c) The induced current on the surface of the superconductor ( xy the plane) can be determined from the boundary condition on the tangential component of B (Eq. 5.76): B=μ0(K×z^). Using the field you get from the image configuration, show that

K=-3mrh2π(r2+h2)52ϕ^

where r is the distance from the origin.

A perfectly conducting spherical shell of radius rotates about the z axis with angular velocity ω, in a uniform magnetic field B=B0Z^. Calculate the emf developed between the “north pole” and the equator. Answer:localid="1658295408106" [12B0ωα2].

A long solenoid, of radius a, is driven by an alternating current, so that the field inside is sinusoidal:Bt=B0cosωtz^. A circular loop of wire, of radius a/2 and resistance R , is placed inside the solenoid, and coaxial with it. Find the current induced in the loop, as a function of time.

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