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A solenoid used to produce magnetic fields for research purposes is 2.1 m long, with an inner radius of 28 cm and 1000 turns of wire. When running, the solenoid produced a field of 1.3 T in the center. Given this, how large a current does it carry?

Respuesta :

Answer:

I = 2172.46 A

Explanation:

Given that,

The length of a solenoid, l = 2.1 m

The inner radius of the solenoid, r = 28 cm = 0.28 m

The number of turns in the wire, N = 1000

The magnetic field in the solenoid, B = 1.3 T

We need to find the current carried by it. We know that, the magnetic field in a solenoid is given by :

[tex]B=\mu_o nI\\\\or\\\\B=\mu_o \dfrac{N}{L}I\\\\I=\dfrac{BL}{\mu_o N}[/tex]

Put all the values,

[tex]I=\dfrac{1.3\times 2.1}{4\pi \times 10^{-7}\times 1000}\\\\I=2172.46\ A[/tex]

So, it carry current of 2172.46 A.

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