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Chapter 7: Q7.34P (page 336)

Question: A fat wire, radius a, carries a constant current I , uniformly distributed over its cross section. A narrow gap in the wire, of width w << a, forms a parallel-plate capacitor, as shown in Fig. 7.45. Find the magnetic field in the gap, at a distance s < a from the axis.

Short Answer

Expert verified

Answer

The magnetic field in the gap is B=渭0Is2蟺a2.

Step by step solution

01

Write the given data from the question.

The radius of the wire is a .

The constant current is the wire is I.

The gap between the wire is w<<a.

02

  Step 2: Determine the formulas to calculate the magnetic field in the gap. 

The expression for the current density is given as follows.

J=IA

Here, A is the area of the wire.

03

Calculate the magnetic field in the gap.  

Consider the figure of the wire with the gap.

Calculate the current density.

Substitute蟺a2 for A into equation (1).

Jd=I蟺a2z^

The expression for the magnetic field from the Ampere鈥檚 law is given by,

鈭�B路dl=渭0IdencB2蟺s=渭0IdencB=渭0Idenc2蟺s

Substitute Jd蟺s2 for Idencinto above equation.

B=渭0Jd蟺s22蟺sB=渭0Jds2

SubstituteI蟺a2forJd into above equation.

B=渭0I蟺a2s2B=渭0Is2蟺a2

Hence the magnetic field in the gap isB=渭0Is2蟺a2.

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Most popular questions from this chapter

A square loop of wire, of side a, lies midway between two long wires,3aapart, and in the same plane. (Actually, the long wires are sides of a large rectangular loop, but the short ends are so far away that they can be neglected.) A clockwise current Iin the square loop is gradually increasing: role="math" localid="1658127306545" dldt=k(a constant). Find the emf induced in the big loop. Which way will the induced current flow?

If a magnetic dipole levitating above an infinite superconducting plane (Pro b. 7 .45) is free to rotate, what orientation will it adopt, and how high above the surface will it float?

Prove Alfven's theorem: In a perfectly conducting fluid (say, a gas of free electrons), the magnetic flux through any closed loop moving with the fluid is constant in time. (The magnetic field lines are, as it were, "frozen" into the fluid.)

(a) Use Ohm's law, in the form of Eq. 7.2, together with Faraday's law, to prove that if 蟽=鈭�and is J finite, then

鈭�B鈭�t=鈭�脳(v脳B)

(b) Let S be the surface bounded by the loop (P)at time t , and S'a surface bounded by the loop in its new position (P')at time t+dt (see Fig. 7.58). The change in flux is

诲桅=鈭�S'B(t+dt)鈥�da-鈭�SB(t)鈥�da

Use 鈭�路B=0to show that

鈭�S'B(t+dt)鈥�da+鈭�RB(t+dt)鈥�da=鈭�SB(t+dt)鈥�da

(Where R is the "ribbon" joining P and P' ), and hence that

诲桅=dt鈭�S鈭�B鈭�t路da-鈭�RB(t+dt)鈥�da

(For infinitesimal dt ). Use the method of Sect. 7.1.3 to rewrite the second integral as

dt鈭�P(B脳v)路dI

And invoke Stokes' theorem to conclude that

诲桅dt=鈭�S(鈭�B鈭�t-鈭�脳v脳B)路da

Together with the result in (a), this proves the theorem.

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

B=渭02蟺Is蠒^

The electric field inside the wire is uniform:

E=I蚁蟺a2z^,

Where 蚁is 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)=I蚁z蟺a2 ; (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 蟽(z)on the wire.

[Answer: V=(-Iz蚁/蟺a2) This is a peculiar result, since Es and 蟽(z)are not independent of localid="1658816847863" z鈫�as one would certainly expect for a truly infinite wire.]

Question: A capacitor C has been charged up to potential V0at time t=0, it is connected to a resistor R, and begins to discharge (Fig. 7.5a).

(a) Determine the charge on the capacitor as a function of time,Q(t)What is the current through the resistor,l(t)?

(b) What was the original energy stored in the capacitor (Eq. 2.55)? By integrating Eq. 7.7, confirm that the heat delivered to the resistor is equal to the energy lost by the capacitor.

Now imagine charging up the capacitor, by connecting it (and the resistor) to a battery of voltage localid="1657603967769" V0, at time t = 0 (Fig. 7.5b).

(c) Again, determine localid="1657603955495" Q(t)and l(t).

(d) Find the total energy output of the battery (鈭�Vldt). Determine the heat delivered to the resistor. What is the final energy stored in the capacitor? What fraction of the work done by the battery shows up as energy in the capacitor? [Notice that the answer is independent of R!]
See all solutions

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