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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

The current in a long solenoid is increasing linearly with time, so the flux is proportional t:.蠒=伪tTwo voltmeters are connected to diametrically opposite points (A and B), together with resistors ( R1and R2), as shown in Fig. 7.55. What is the reading on each voltmeter? Assume that these are ideal voltmeters that draw negligible current (they have huge internal resistance), and that a voltmeter register --鈭�abE脳dlbetween the terminals and through the meter. [Answer: V1=伪R1/(R1+R2). Notice that V1鈮�V2, even though they are connected to the same points]

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.]

A square loop, side a , resistance R , lies a distance from an infinite straight wire that carries current l (Fig. 7.29). Now someone cuts the wire, so l drops to zero. In what direction does the induced current in the square loop flow, and what total charge passes a given point in the loop during the time this current flows? If you don't like the scissors model, turn the current down gradually:

I(t)={(1-t)I0for0鈮�t鈮�1/afort>/a

A toroidal coil has a rectangular cross section, with inner radius a , outer radius a+w, and height h . It carries a total of N tightly wound turns, and the current is increasing at a constant rate (dl/dt=k). If w and h are both much less than a , find the electric field at a point z above the center of the toroid. [Hint: Exploit the analogy between Faraday fields and magnetostatic fields, and refer to Ex. 5.6.]

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.
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