# When the current in a toroidal solenoid is changing at a rate of 0.0260 A/s, the magnitude of the induced emf is 12.6 mV. When the current e

Question

When the current in a toroidal solenoid is changing at a rate of 0.0260 A/s, the magnitude of the induced emf is 12.6 mV. When the current equals 1.40 A, the average flux through each turn of the solenoid is 0.00285 Wb. How many turns does the solenoid have?

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1 year 2021-08-11T14:22:45+00:00 1 Answers 4 views 0

N= 238 turns

Explanation:

The induced Emf that goes through a solenoid can be calculated using the below formula;

Where ξ=induced Emf

L= self inductance

I= current

ξ= L|dⁱ/dt|

Making L which is the self inductance subject of formula we have

L=ξ/[|dⁱ|*|dt|]

The current here is changing at the rate of

.0260 A/s

L=NΦB/i

N=ξ/Φ|di|*|dt|

Magnitude of the induced Emf given= 12.6mV then if we convert to volt we have 12.6×10⁻³ V

The current I = 1.40A

Magnitude flux through the flux=/0.00285 Wb

Then if we substitute all this Value to equation above we have

N=(12.6×10⁻³ V×1.40A)/(0.00285 Wb×0.0260 A/s)

N=238turn

Therefore, there are 238turns in the solenoid