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A long solenoid that has 1 080 turns uniformly distributed over a length of 0.420 m produces a magnetic field of magnitude 1.00 10-4 T at its center. What current is required in the windings for that to occur?

1 Answer

6 votes

Answer:

Current, I = 0.073 A

Step-by-step explanation:

It is given that,

Number of turns in a long solenoid is 1080

Length of the solenoid is 0.420 m

It produces a magnetic field of
10^(-4)\ T at its center.

We need to find the current is required in the winding for that to occur. The magnetic field at the center of the solenoid is given by :


B=\mu_0 NI

I is current


I=(B)/(\mu_o N)\\\\I=(10^(-4))/(4\pi * 10^(-7)* 1080)\\\\I=0.073\ A

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