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Two concentric metal spherical shells, of radius a and b, respectively, are separated by weakly conducting material of conductivity(Fig. 7 .4a).

(a) If they are maintained at a potential difference V, what current flows from one to the other?

(b) What is the resistance between the shells?

(c) Notice that if b>>a the outer radius (b) is irrelevant. How do you account for that? Exploit this observation to determine the current flowing between two metal spheres, each of radius a, immersed deep in the sea and held quite far apart (Fig. 7 .4b ), if the potential difference between them is V. (This arrangement can be used to measure the conductivity of sea water.)

Short Answer

Expert verified

(a) The expression for the current isI=4(VaVb)(1a1b) .

(b) The resistance between the shells is14(1a1b) .

(c) The expression for the current between the two sphere is2Va .

Step by step solution

01

Determine the formula for the electric field as 

Consider the formula for the electric field

E=140Qr2

Here0, is the permittivity of the free space,Q is the charge andr is the distance between the sphere.

Consider the expression for the current is

I=VR

02

(a) Determine the value of the current flowing

Determine the electric filed between concentric metal spheres.

E=140Qr2

If the voltage potential difference is Vin the concentric spheres having radius aand b.

Write the expression for the voltage difference as

VaVb=baQ401r2dr=Q40ba1r2dr=Q40(1a1b) 鈥.. (1)

Consider the formula for the electric current in terms of the electric current density is

I=Eda=Q0

From equation (1) rewrite the expression for current as

.I=40(VaVb)0(1a1b)I=4(VaVb)(1a1b)

Therefore, the expression for the current isI=4(VaVb)(1a1b) .

03

(b) Determine the resistance between the shells

Consider the formula for the resistance as

R=VI

Rewrite the expression for the resistance in terms of the voltage difference as

R=VaVb4(VaVb)(1a1b)=14(1a1b)

04

(c) Determine the current between the two spheres

Consider that b>>>a here, on negating athe sphere feel current by the sphere b on the basis of the difference between both the sphere. The expression is鈥

R=14a

Since, the resistance is due to the inner sphere, the successive shells have less contribution in the current because of the small cross sectional area.

Write the expression for the two submerged sphere as

R=24a=12a

From the general expression for the resistance solve as

R=VII=V12aI=2Va

Therefore, the expression for the current between the two sphere is 2Va.

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Most popular questions from this chapter

A circular wire loop (radiusr, resistanceR) encloses a region of uniform magnetic field,B, perpendicular to its plane. The field (occupying the shaded region in Fig. 7.56) increases linearly with timeB=t. An ideal voltmeter (infinite internal resistance) is connected between pointsPandQ.

(a) What is the current in the loop?

(b) What does the voltmeter read? [Answer: r2/2]

A long solenoid with radius a and n turns per unit length carries a time-dependent currentl(t) in the^ 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.

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 circular wire loop (radius r , resistance R ) encloses a region of uniform magnetic field, B , perpendicular to its plane. The field (occupying the shaded region in Fig. 7.56) increases linearly with time(B=t)An ideal voltmeter (infinite internal resistance) is connected between points P and Q.

(a) What is the current in the loop?

(b) What does the voltmeter read? Answer:[r2/2]

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 =and is J finite, then

Bt=(vB)

(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=S'B(t+dt)da-SB(t)da

Use B=0to show that

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

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

诲桅=dtSBtda-RB(t+dt)da

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

dtP(Bv)dI

And invoke Stokes' theorem to conclude that

诲桅dt=S(Bt-vB)da

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

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