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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=μ0Is2πa2.

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.

Substituteπa2 for A into equation (1).

Jd=Iπa2z^

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

∮B·dl=μ0IdencB2πs=μ0IdencB=μ0Idenc2πs

Substitute Jdπs2 for Idencinto above equation.

B=μ0Jdπs22πsB=μ0Jds2

SubstituteIπa2forJd into above equation.

B=μ0Iπa2s2B=μ0Is2πa2

Hence the magnetic field in the gap isB=μ0Is2πa2.

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Most popular questions from this chapter

A rectangular loop of wire is situated so that one end (height h) is between the plates of a parallel-plate capacitor (Fig. 7.9), oriented parallel to the field E. The other end is way outside, where the field is essentially zero. What is the emf in this loop? If the total resistance is R, what current flows? Explain. [Warning: This is a trick question, so be careful; if you have invented a perpetual motion machine, there's probably something wrong with it.]

A battery of emf εand internal resistance r is hooked up to a variable "load" resistance,R . If you want to deliver the maximum possible power to the load, what resistance R should you choose? (You can't change e and R , of course.)

A square loop is cut out of a thick sheet of aluminum. It is then placed so that the top portion is in a uniform magnetic field , B and is allowed to fall under gravity (Fig. 7 .20). (In the diagram, shading indicates the field region; points into the page.) If the magnetic field is 1 T (a pretty standard laboratory field), find the terminal velocity of the loop (in m/s ). Find the velocity of the loop as a function of time. How long does it take (in seconds) to reach, say, 90% of the terminal velocity? What would happen if you cut a tiny slit in the ring, breaking the circuit? [Note: The dimensions of the loop cancel out; determine the actual numbers, in the units indicated.]

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

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 (∫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!]
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