Answered

A 28 cm long solenoid 1.25 cm in diameter is to produce a field of 0.353 T at its center. How much current should the solenoid carry if it has 97 turns of the wire?

Answer :

Answer:

810.9 A

Explanation:

Length of the solenoid = 28 cm =0.28 m

Magnetic field B =0.353 T

Number of turns N =97

We know that the magnetic field due to solenoid [tex]B=\mu _0ni[/tex] where [tex]n=\frac{N}{l}[/tex]

So [tex]0.353=\frac{4\pi \times 10^{-7}\times 97\times i}{0.28}[/tex]

Current i =810.9 A

So the current in the solenoid will be 810.9 A

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