Where is Btnonzero 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 Btis 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, withBE andμ0Iencdt 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 termBt is nonzero along the left and right edges of the shaded rectangle.

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Most popular questions from this chapter

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.]

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]

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.

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?

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

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