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Chapter 30: Q44P (page 899)

A12 Hinductor carries a current of2.0 A. At what rate must the current be changed to produce a60 Vemf in the inductor?

Short Answer

Expert verified

didt=-5.0A/s

Step by step solution

01

Given

  1. Inductance of inductor, L=12.0H
  2. Current through inductor,i=2.0A
  3. Emf induced, ε=60V
02

Understanding the concept

We can use Faraday’s law of electromagnetic induction and the definition of inductance to find the rate of change of current through the inductor.

Formula:

ε=-dϕBdt

ϕB=Li

03

Calculate at what rate must the current be changed to produce a emf in the inductor

We have

ε=-dϕBdt=-ddtLi

As the inductance is fixed for a given inductor, then

ε=-dϕBdt=-Ldidt

So using the above equation we get

didt=-εL=6012.0=- 5.0A/s

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Most popular questions from this chapter

The switch in the circuit of Fig. 30-15 has been closed on a for a very long time when it is then thrown to b. The resulting current through the inductor is indicated in Fig. 30-28 for four sets of values for the resistance R and inductance L: (1) R=R0, (2) 2R0=R, (3) R0and2L0 , (4) 2R0and2L0. Which set goes with which curve?

Figure shows a copper strip of width W = 16.0 cmthat has been bent to form a shape that consists of a tube of radius R = 1.8 cmplus two parallel flat extensions. Current i = 35 mAis distributed uniformly across the width so that the tube is effectively a one-turn solenoid. Assume that the magnetic field outside the tube is negligible and the field inside the tube is uniform. (a) What is the magnetic field magnitude inside the tube? (b) What is the inductance of the tube (excluding the flat extensions)?

In Figure (a), the inductor has 25 turns and the ideal battery has an emf of 16 V. Figure (b) gives the magnetic fluxφ through each turn versus the current i through the inductor. The vertical axis scale is set by φs=4.0×10-4Tm2, and the horizontal axis scale is set by is=2.00AIf switch S is closed at time t = 0, at what ratedidtwill the current be changing att=1.5τL?

Question: At a certain place, Earth’s magnetic field has magnitudeB=0.590gaussand is inclined downward at an angle of 70.0°to the horizontal. A flat horizontal circular coil of wire with a radius of 10.0 cmhas 1000 turnsand a total resistance of85.0Ω. It is connected in series to a meter with140Ωresistance. The coil is flipped through a half-revolution about a diameter, so that it is again horizontal. How much charge flows through the meter during the flip?

Two coils connected as shown in Figure separately have inductances L1 and L2. Their mutual inductance is M. (a) Show that this combination can be replaced by a single coil of equivalent inductance given by

Leq=L1+L2+2M

(b) How could the coils in Figure be reconnected to yield an equivalent inductance of

Leq=L1+L2-2M

(This problem is an extension of Problem 47, but the requirement that the coils be far apart has been removed.)
See all solutions

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