Physics, asked by empiarp, 1 month ago

A current i = 1.5 A is flowing through a long solenoid of diameter 3.2 cm, having 220 turns per cm. At its centre, a 130 tum closely packed coll of diameter 2.1 cm is placed such that the coil is is coaxial with the long solenoid. The current in the solenoid is reduced to zero at a steady in 25 ms. What is the magnitude of emf induced in the coil while the current in the solenoid is changing?

Answers

Answered by XxIndianpilotxX
1

Answer:

Field through inner solenoid

n=22000turns/m,i=1.5A

B=μ

0

ni=4π×10

−7

×22000×1.5

=0.0415T

emf induced ϵ

0

=

dt

dQ

B

=

25×10

−3

130×0.0415×π×(0.0105)

2

=0.0748v

Explanation:

I HOPE THAT IT'S HELPFUL FOR YOU.

Answered by mikasa15
4

Hey there !

• Initially magnetic flux passing through the coil ( single turn ) ,

Φ = B.A cos 0°

= µ0 n l x πd^2 /4

= 4π x 10^-7 x 1.5 x 22000 x 3.464 x

10^-4

= 1.44 x 10^-5 Wb.

• Finally the flux becomes zero because the current reduces to zero.

Thus , e = |∆Φ| / ∆t

= 1.44x 10^-5 / 25 x 10^-3

= 5.76 x 10^-4 V

• Hence , total emf = N x e

= 130 x 5.76 x 10^-4

= 75 mV

# Hope it helps!

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