(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?

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

two charges are fixed in place on the x-axis as shown. the charge at x = 0 is negative and has magnitude q = 2 μcoulombs (1 μcoulomb = 10-6 coulombs). the charge at x = d, (d = +12 cm), is positive and has magnitude q = 4 μcoulombs. find a point x = xo along the x-axis at which the electric field is zero.

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