A coil is connected in series with a 10.0kresistor. An ideal 50.0 Vbattery is applied across the two devices, and the current reaches a value of 2.00 mAafter 5.00 ms.(a) Find the inductance of the coil. (b) How much energy is stored in the coil at this same moment?

Short Answer

Expert verified

a) L=97.9H

b)Ub=0.196mJ

Step by step solution

01

Given

R=10k=10×103E=50.0VI=2.00mA=2.00×10-3At=5.00ms=5×10-3s

02

Understanding the concept

Here we have to use the formula for current through RL circuit to find inductance. Then by using formula for energy stored by inductor’s magnetic field, we can calculate the energy stored in the coil.

Formula:

i=ER1-etLtUb=0.5Li2

03

(a) Find the inductance of the coil

Here, the rise of current for RL circuit is as follows:

i=ER1-e-tLti=ER1-e-t×RL2.00×10-3=5010×1031-e-5×10-3×10×103L0.4=1-e50-L0.6=e-50/LL=-50In0.6L=97.9H

04

(b) Calculate how much energy is stored in the coil at this same moment

The inductor’s magnetic field stores energy and is given by

Ub=0.5Li2Ub=0.5×97.9×(2.00×10-3)2

Ub=0.196mJ

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Most popular questions from this chapter

Question: An electric generator contains a coil of 100 turnsof wire, each forming a rectangular loop 50.0cm to 30.0cm. The coil is placed entirely in a uniform magnetic field with magnitude B = 3.50 Tand withBinitially perpendicular to the coil’s plane. What is the maximum value of the emf produced when the coil is spun at 1000 rev/min about an axis perpendicular toB?

How long would it take, following the removal of the battery, for the potential difference across the resistor in an RL circuit (with L = 2.00H, R = 3.00) to decay to 10.0% of its initial value?

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)?

The current in an RL circuit builds up to one-third of its steady-state value in 5.00 s . Find the inductive time constant.

:Inductors in parallel. Two inductors L1 and L2 are connected in parallel and separated by a large distance so that the magnetic field of one cannot affect the other. (a)Show that the equivalent inductance is given by

1Leq=1L2+1L2

(Hint: Review the derivations for resistors in parallel and capacitors in parallel. Which is similar here?) (b) What is the generalization of (a) for N inductors in parallel?

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