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3 years ago

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(a) Calculate the self-inductance of a 42.2 cm long, 10.0 cm diameter solenoid having 1000 loops. (b) How fast can it be turned

off if the induced emf cannot exceed 3.00 V and a current of 18.6 A flows through this inductor?

1 answer:

kaheart [24]3 years ago

6 0

Answer:

self-inductance is 23.4 mH

0.145 seconds it be turned off

Explanation:

given data

length L = 42.2 cm

diameter d = 10 cm

no of loop N = 1000 loops

induced emf = 3 V

current I = 18.6 A

to find out

self-inductance and How fast it turned off emf

solution

we apply here formula of self-inductance that is

self-inductance = u×N²×A / L .......................1

put here value

self-inductance = 4π×1000²×(π×r² / 0.422)

self-inductance = 4π×1000²×(π×0.05² / 0.422)

self-inductance = 0.0234 H

so self-inductance is 23.4 mH

and

we know emf = Ldi /dt

so time = 0.0234 × 18.6

time = 0.145 seconds

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See the last 7 answers in the attached document.

<span class="sg-text sg-text--link sg-text--bold sg-text--link-disabled sg-text--blue-dark"> docx </span>

<span class="sg-text sg-text--link sg-text--bold sg-text--link-disabled sg-text--blue-dark"> pdf </span>

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3 years ago