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Suppose

E(r,t)=14πε0qr2θ(rυt)r^; B(r,t)=0

(The theta function is defined in Prob. 1.46b). Show that these fields satisfy all of Maxwell's equations, and determine ρ and J. Describe the physical situation that gives rise to these fields.

Short Answer

Expert verified

The value offirst Maxwell’s equation for the given functions of B and Eare.ρ=qδ3(r)θ(t)+q4πr2δ(υtr)

The value of second Maxwell’s equation for the given functions of B and E areB=0.

The value of Third Maxwell’s equation for the given functions of B and E are×E=0 .

The value of fourth Maxwell’s equation for the given functions of B and Eare J=(q4πr2)(δ(υtr))r^.

Step by step solution

01

Write the given data from the question.

Consider thegiven electrical field isE(r,t)=14πε0qr2θ(υtr)r^.

Consider thegiven magnetic field isB(r,t)=0.

Consider the Maxwell’s equations are given by

E=ρε0B=0×E=Bt×B=μ0J+μ0ε0Et

02

Determine the formulaof Maxwell’s equation for the given functions of  and

Write the formula of first Maxwell’s equation for the given functions of Band E.

E=ρε0…… (1)

Here, ρ is a charge and ε0 is absolute permittivity.

Write the formula of second Maxwell’s equation for the given functions of BandE.

B

Here, is derivative and B is magnetic field.

Write the formula of third Maxwell’s equation for the given functions of BandE.

×E…… (2)

Here, is derivative, B is electrical field.

Write the formula of fourth Maxwell’s equation for the given functions of andE.

×B=μ0J+μ0ε0Εt…… (3)

Here,μ0 is permeability and J is current density, E is permittivity and is electric field.

03

Step 3:Determine thevalue Maxwell’s equation for the given functions of  B and E

Determine the value of first Maxwell’s equation for the given functions of Band.E

Substitute θ(υtr) for and (14πε0qr2r^)14πε0qr2r^|θ(υtr)| into equation (1).

E=θ(υtr)(14πε0qr2r^)14πε0qr2r^|θ(υtr)|=qε0δ3(r)θ(υtr)14πε0qr2(r^r^)rθ(υtr)

From problem 1.45, we are given the next δ3(r)θ(υtr)δ3(r)θ(t)and rθ(υtr)δ(υtr). So, plug these expressions into equation (1) to get.

E=qε0δ3(r)θ(υtr)14πε0qr2(r^r^)rθ(υtr)ρ=ε0Eqδ3(r)θ(t)+q4πr2δ(υtr)

Determine the value of second Maxwell’s equation for the given functions of Band.E

Plug the expression ofB, so we get it by

Substitute (0) for and (0)=0

Determine the value of third Maxwell’s equation for the given functions of Band E.

Given that the electric field is independent of θ and ϕ the vector product will be zero.

Substitute ϕ for Btinto equation (2).

×E=0

Determine the value of fourth Maxwell’s equation for the given functions of BandE.

To obtain the current density, plug in the formula for B and E as follows:

Substitute (14πε0qr2θ(υtr))r^ for E into equation (3).

×0=μ0J+μ0ε0t(14πε0qr2θ(υtr))r^0=J+ε0t(14πε0qr2θ(υtr))r^J=ε0t(14πε0qr2θ(υtr))r^=ε0(14πε0)qr2t(θ(υtr))r^

Solve further as

J=(q4πr2)(δ(υtr))r^.

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Most popular questions from this chapter

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.]

Problem 7.61 The magnetic field of an infinite straight wire carrying a steady current I can be obtained from the displacement current term in the Ampere/Maxwell law, as follows: Picture the current as consisting of a uniform line charge λmoving along the z axis at speed v (so that I=λv), with a tiny gap of length E , which reaches the origin at time t=0. In the next instant (up to t=E/v) there is no real current passing through a circular Amperian loop in the xy plane, but there is a displacement current, due to the "missing" charge in the gap.

(a) Use Coulomb's law to calculate the z component of the electric field, for points in the xy plane a distances from the origin, due to a segment of wire with uniform density -λ . extending from toz1=vt-Etoz2=vt .

(b) Determine the flux of this electric field through a circle of radius a in the xy plane.

(c) Find the displacement current through this circle. Show thatId is equal to I , in the limit as the gap width (E)goes to zero.35

A long solenoid with radius a and n turns per unit length carries a time-dependent currentl(t) in theϕ^ direction. Find the electric field (magnitude and direction) at a distance s from the axis (both inside and outside the solenoid), in the quasistatic approximation.

Imagine a uniform magnetic field, pointing in the zdirection and filling all space (B=B0z). A positive charge is at rest, at the origin. Now somebody turns off the magnetic field, thereby inducing an electric field. In what direction does the charge move?

In a perfect conductor, the conductivity is infinite, so E=0(Eq. 7.3), and any net charge resides on the surface (just as it does for an imperfect conductor, in electrostatics).

(a) Show that the magnetic field is constant (Bt=0), inside a perfect conductor.

(b) Show that the magnetic flux through a perfectly conducting loop is constant.

A superconductor is a perfect conductor with the additional property that the (constant) B inside is in fact zero. (This "flux exclusion" is known as the Meissner effect.)

(c) Show that the current in a superconductor is confined to the surface.

(d) Superconductivity is lost above a certain critical temperature (Tc), which varies from one material to another. Suppose you had a sphere (radius ) above its critical temperature, and you held it in a uniform magnetic field B0z^while cooling it below Tc. Find the induced surface current density K, as a function of the polar angleθ.

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