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Where is ∂B∂tnonzero in Figure 7.21(b)? Exploit the analogy between Faraday's law and Ampere's law to sketch (qualitatively) the electric field.

Short Answer

Expert verified

The term ∂B∂tis nonzero along the left and right edges of the shaded rectangle.

Step by step solution

01

Define Faraday's Law.

Faraday's law is the most fundamental law in Electromagnetic induction, which depicts that whenever a conductor is placed in the magnetic field, which is variable in nature, then an EMF will be induced in the conductor if the conductor has been short circuited, then current will flow in the conductor.

02

Determine the nonzero position of ∂B∂t

The figure concludes that the (inward) flux through the strip on the left isincreasing. On the contrary, the (inward) flux passing through the strip on the right decreases, similar to the two current sheets under Ampere's circuital law.

Now, withB→E andμ0Ienc→−»åÏ•dt from equation 7.19, the significance of the negative sign is that the current has been flown out to one on the left, so its field is counter clockwise, and the one on the right is like a current flowing in, so it field is clockwise.

The diagram of the above description is shown below:

Therefore, the term∂B∂t is nonzero along the left and right edges of the shaded rectangle.

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Most popular questions from this chapter

A familiar demonstration of superconductivity (Prob. 7.44) is the levitation of a magnet over a piece of superconducting material. This phenomenon can be analyzed using the method of images. Treat the magnet as a perfect dipole , m a height z above the origin (and constrained to point in the z direction), and pretend that the superconductor occupies the entire half-space below the xy plane. Because of the Meissner effect, B = 0 for Z≤0, and since B is divergenceless, the normal ( z) component is continuous, so Bz=0just above the surface. This boundary condition is met by the image configuration in which an identical dipole is placed at - z , as a stand-in for the superconductor; the two arrangements therefore produce the same magnetic field in the region z>0.

(a) Which way should the image dipole point (+ z or -z)?

(b) Find the force on the magnet due to the induced currents in the superconductor (which is to say, the force due to the image dipole). Set it equal to Mg (where M is the mass of the magnet) to determine the height h at which the magnet will "float." [Hint: Refer to Prob. 6.3.]

(c) The induced current on the surface of the superconductor ( xy the plane) can be determined from the boundary condition on the tangential component of B (Eq. 5.76): B=μ0(K×z^). Using the field you get from the image configuration, show that

K=-3mrh2Ï€(r2+h2)52Ï•^

where r is the distance from the origin.

Refer to Prob. 7.16, to which the correct answer was

E(s,t)=μ0I0Ӭ2ττsin(Ӭt)In(as)z^

(a) Find the displacement current density Jd·

(b) Integrate it to get the total displacement current,

Id=∫Jd.da

Compare Id and I. (What's their ratio?) If the outer cylinder were, say, 2 mm in diameter, how high would the frequency have to be, forId to be 1% of I ? [This problem is designed to indicate why Faraday never discovered displacement currents, and why it is ordinarily safe to ignore them unless the frequency is extremely high.]

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?

Question:A long cable carries current in one direction uniformly distributed over its (circular) cross section. The current returns along the surface (there is a very thin insulating sheath separating the currents). Find the self-inductance per unit length.

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