| Grade Summary Submissions 12 Begin Date: 10/7/2017 8:15:00 PM -- Due Date: 11/
ID: 1792761 • Letter: #
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Grade Summary
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12 Begin Date: 10/7/2017 8:15:00 PM -- Due Date: 11/2/2017 7:00:00 PM End Date: 11/2/2017 7:00:00 PM (25%) Problem 3: Suppose a 52.5 cm long, 11.5 cm diameter solenoid has 1000 loops. 33% Part (a) Calculate the self-inductance of it in mH. L =|
sin() cos() tan() cotan() asin() acos() atan() acotan() sinh() cosh() tanh() cotanh() Degrees Radians ( ) 7 8 9 HOME ^ ^ 4 5 6 / * 1 2 3 + - 0 . END () BACKSPACE DEL CLEARGrade Summary
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Attempts remaining: 20 (0% per attempt) detailed view . 33% Part (b) How much energy is stored in this inductor when 17.5 A of current flows through it? Give your answer in J.33% Part (c) How fast can it be turned off (in s) if the average induced emf cannot exceed 2.95 V?
Explanation / Answer
a) A = pi d2 / 4 = pi * 0.1152 / 4 = 0.0104 m2
L = mu0 N2 A / l = (4 pi * 10-7 * 10002 * 0.0104) / (0.525)
= 24.89 mH
b) Energy stored = 1/2 L i2
= 1/2 * 24.89 * 10-3 * 17.52
energy stored = 3.81 J
c) dt = - L * di / 2.95 = -24.89 * 10-3 * (0 - 17.5) / 2.95
= 0.148 s
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