Problem 3: An inductor with L = 8.5 H is in a closed circuit with I = 0.55 A (see thediagram). When the switch is opened, the current goes to zero in At = 0.14 s.%3D%3D%3DLRandomized VariablesllL = 8.5 HI = 0.55 AAt = 0.14 s%3DI%3DOtheexpertta.corPart (a) Express the magnitude of the average induced emf on the inductor, ɛave, in terms of L, I and At.Expression :%3DEave =Select from the variables below to write your expression. Note that all variables may not be required.a, ß, At, 0, d, g, h, i, I, j, k, L, m, n, PPart (b) Calculate the numerical value of ɛave in volts.Numeric : A numeric value is expected and not an expression.%3DEavettps://ush22md.theexpertta.com/Common/ViewAssignmentDetails.aspx

Answered: Image /qna-images/answer/726a0004-e7bd-471b-bb40-47d9e48e77f3.jpg

Problem 3: An inductor with L 8.5 H is in a closed circuit with I 0.55 A (see the diagram). When the switch is opened, the current goes to zero in At = 0.14 s. Randomized Variables L = 8.5 H I = 0.55 A At = 0.14 s

Problem 3: An inductor with L 8.5 H is in a closed circuit with I 0.55 A (see the diagram). When the switch is opened, the current goes to zero in At = 0.14 s. Randomized Variables L = 8.5 H I = 0.55 A At = 0.14 s

Physics for Scientists and Engineers, Technology Update (No access codes included)

9th Edition

ISBN:9781305116399

Author:Raymond A. Serway, John W. Jewett

Publisher:Raymond A. Serway, John W. Jewett

Chapter32: Inductance

Section: Chapter Questions

Problem 32.27P

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Concept explainers

Electromagnets

The word "electromagnet" is composed of two words, ‘electro’ and ‘magnet’. It is a type of magnet in which an electric current helps in the generation of a magnetic field. Electromagnets usually consist of two major parts, wire and a coil around which the wire is wound. This is called a solenoid. When an electric current is allowed to pass through the coiled wire, the electric current produces a magnetic field. This magnetic field is present only when current is passing through the wire.

Question

Problem 3: An inductor with L = 8.5 H is in a closed circuit with I = 0.55 A (see the
diagram). When the switch is opened, the current goes to zero in At = 0.14 s.
%3D
%3D
%3D
L
Randomized Variables
ll
L = 8.5 H
I = 0.55 A
At = 0.14 s
%3D
I
%3D
Otheexpertta.cor
Part (a) Express the magnitude of the average induced emf on the inductor, ɛave, in terms of L, I and At.
Expression :
%3D
Eave =
Select from the variables below to write your expression. Note that all variables may not be required.
a, ß, At, 0, d, g, h, i, I, j, k, L, m, n, P
Part (b) Calculate the numerical value of ɛave in volts.
Numeric : A numeric value is expected and not an expression.
%3D
Eave
ttps://ush22md.theexpertta.com/Common/ViewAssignmentDetails.aspx

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