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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?

Short Answer

Expert verified

The direction of the charge is indetermined.

Step by step solution

01

Write the given data from the question.

The uniform magnetic field B=B0z,

The initial magnetic field isB

The positive charge is at rest.

02

Determine the direction of the charge move.

Let assume the charge isqat the rest, and Eis the electric field.

The force experienced by the charges at the rest is given by,

F=qE

The curl of the electric field is given by,

×E=BC^tZ

Since the boundary condition is not mentioned therefore, it is difficult to move the electric field.

The electric field attain the zero value at infinity. Hence the direction of the charge is in determined due to no boundary condition.

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Most popular questions from this chapter

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 σ=and is J finite, then

Bt=×(v×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

=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

=dtSBt·da-RB(t+dt)da

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

dtP(B×v)·dI

And invoke Stokes' theorem to conclude that

dt=S(Bt-×v×B)·da

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

Try to compute the self-inductance of the "hairpin" loop shown in Fig. 7.38. (Neglect the contribution from the ends; most of the flux comes from the long straight section.) You'll run into a snag that is characteristic of many self-inductance calculations. To get a definite answer, assume the wire has a tiny radius, and ignore any flux through the wire itself.

Question: Assuming that "Coulomb's law" for magnetic charges ( qm) reads

F=μ04πqm1qm2r2r^

Work out the force law for a monopole moving with velocity through electric and magnetic fields E and B.

A long solenoid of radius a, carrying n turns per unit length, is looped by a wire with resistance R, as shown in Fig. 7.28.

(a) If the current in the solenoid is increasing at a constant rate (dl/dt=k),, what current flows in the loop, and which way (left or right) does it pass through the resistor?

(b) If the currentlin the solenoid is constant but the solenoid is pulled out of the loop (toward the left, to a place far from the loop), what total charge passes through the resistor?

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=αt. 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: αr2/2]

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