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4 years ago

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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 t

he 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?

1 answer:

andrew11 [14]4 years ago

4 0

Answer:

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

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