Question
Asked Nov 30, 2019
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An air-core solenoid with 66 turns is 8.00 cm long and has a diameter of 1.20 cm. When the solenoid carries a current of 0.740 A, how much energy is stored in its magnetic field in µJ?

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Expert Answer

Step 1

Self-inductance for an air cored solenoid is given as:

N2A
L-
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N2A L-

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Step 2

Calculating the inductance produced in the coil.

N2A
L=
1
(66x66)x(0.012mx0.012m)
4
L-(4tx 10
0.08m
L-77.3x10 H
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N2A L= 1 (66x66)x(0.012mx0.012m) 4 L-(4tx 10 0.08m L-77.3x10 H

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Step 3

Energy stored in its magneti...

E=-I
E(77.3x10H)(0.74A)
E-21.16x10 J
E-21.16x10 J
E-2.116uJ
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E=-I E(77.3x10H)(0.74A) E-21.16x10 J E-21.16x10 J E-2.116uJ

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Current Electricity

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