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The magnetic field of a long, straight, and closely-wound solenoid, inside the solenoid at a point near the center, is 0.645 T. What is the magnitude of the magnetic field at this point if the current in the solenoid is increased by a factor of 3.00

Respuesta :

Answer:

B'=1.935 T      

Explanation:

Given that

magnetic field ,B= 0.645 T

We know that magnetic filed in the solenoid is given as

[tex]B=\mu _0 n\ I [/tex]

I=Current

n=Number of turn per unit length

μ0 =magnetic permeability

Now when the current increased by 3 factors

I'=3 I

Then the magnetic filed

[tex]B'=\mu _0 n\ I'[/tex]

[tex]B'=\mu _0 n\ (3I)[/tex]

B'=3 B

That is why

B' = 3 x 0.645 T

B'=1.935 T

Therefore the new magnetic filed will be 1.935 T.

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