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A tightly wound solenoid is 15 cm long, has 350 turns, and carries a current of 4.0 A. If you ignore end effects, you will find that the value of app at the center of the solenoid when there is no core is approximately

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onyebuchinnaji

Answer:

The magnetic field at the center of the solenoid is approximately  0.0117 T

Explanation:

Given;

length of the solenoid, L = 15 cm = 0.15 m

number of turns of the solenoid, N = 350 turns

current in the solenoid, I = 4.0 A

The magnetic field at the center of the solenoid is given by;

[tex]B = \mu_o (\frac{N}{L} )I\\\\B = (4 \pi *10^{-7})(\frac{350}{0.15} )(4.0)\\\\B = 0.0117 \ T[/tex]

Therefore, the magnetic field at the center of the solenoid is approximately  0.0117 T.

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