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Yakvenalex [24]
2 years ago
3

A 10-turn ideal solenoid has an inductance of 3.5 mH. When the solenoid carries a current of 2.0 A the magnetic flux through eac

h turn is:
Physics
1 answer:
valkas [14]2 years ago
5 0

Answer:

7 * 10^-4Wb

Explanation:

The magnetic flux is the product of the inductance and current

Magnetic flux = LI

L is the inductance = 3.5mH

I is the current = 2.0A

Number of turns N = 2A

Since for a 10-turn solenoid, L = 3.5mH

For a one-turn solenoid, L = 3.5/10 = 0.35mH

magnetic flux = 0.35mH * 2

magnetic flux = 0.35 * 10^-3 * 2

magnetic flux = 0.7* 10^-3

Magnetic flux = 7 * 10^-4Wb

Hence the magnetic flux through each turn is 7 * 10^-4Wb

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