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Chapter 30: Q68P (page 900)

A toroidal inductor with an inductance of 9.0.mH encloses a volume of 0.0200m3. If the average energy density in the toroid is70.0J/m3, what is the current through the inductor?

Short Answer

Expert verified

5.58 A

Step by step solution

01

Given

i) Inductance of inductorL=90.0mH=90.0×10-3H

ii) Volume of toroidV=0.0200m3

ii) Average densityuB=70.0J/m3l

02

Understanding the concept

We use the formula of total energy stored for volumein the magnetic field into the formula of energy stored in the inductor’s magnetic field to find the current through the inductor.

Formula:

UB=12Li2UB=uBV

03

Calculate the current through the inductor

The magnetic field stored in the toroid is given by

UB=12Li2

Butthe total energy stored for volume V in the magnetic field is given by

UB=uBV

Therefore,

12Li2=uBVi=2uBVLi=270.0Jm30.0200m390.0×10-3Hi=5.58A

Therefore, the current through the inductor is 5.58 A.

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Most popular questions from this chapter

Suppose the emf of the battery in the circuit shown in Figure varies with time t so that the current is given by i(t) = 3.0 +5.0 t , where i is in amperes and t is in seconds. Take R=4.0ΩandL=5.0H, and find an expression for the battery emf as a function of t. (Hint: Apply the loop rule.)

Figure 30-31 shows three situations in which a wire loop lies partially in a magnetic field.The magnitude of the field is either increasing or decreasing, as indicated. In each situation, a battery is part of the loop. In which situations are the induced emf and the battery emf in the same direction along the loop?

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