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ahrayia [7]
3 years ago
6

A solenoid used to produce magnetic fields for research purposes is 2.2 mm long, with an inner radius of 30 cmcm and 1200 turns

of wire. When running, the solenoid produced a field of 1.4 TT in the center. Given this, how large a current does it carry?
Physics
1 answer:
serious [3.7K]3 years ago
7 0

Answer:

The current is  I  =  2042\ A

Explanation:

From the question we are told that

    The length of the solenoid is  l  =  2.2 \ m

    The  radius is  r_i  =  30 \ cm  =  0.30  \ m

    The number of turn is N  = 1200 \ turns

    The  magnetic field is  B =  1.4 \ T

The  magnetic field produced  is mathematically represented as

         B  =  \frac{\mu_o *  N  *  I }{l }

making I the subject

       I  =  \frac{B *  l}{\mu_o  *  N }

Where  \mu_o is the permeability of free space with values \mu_o  = 4\pi *10^{-7} N/A^2

 substituting values

        I  =  \frac{1.4  *   2.2 }{4\pi *10^{-7}  *  1200 }

        I  =  2042\ A

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