Question

A rod of magnetic material, 0.5 m in length has a coil of 200 turns wound over it uniformly. If a current of 2 ampere Is sent through it, calculate (a) the magnetic field H, (b) the Intensity of magnetization M, (c) the magnetic induction B and (d) the relative permeability μr of the material. Give x,,, = 6 X 10-n​

Answer 1

Answer:

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Answer 2

Answer:

The answer is  B=25.12T.

Explanation:

Given that,

Relative permeability, μr=4000

Carried current, I=5A

Number of turns, n=1000

Therefore,

The magnetic intensity,

H=nI

H=1000×5

H=5000Am−1

Now, the magnetic field inside the solenoid is

B=μH…. (I)

We know that,

  μ=μ0μr

μ=4π×10−7×4000

Now put the value of μ in equation (I)

B=μH

B=4π×10−7×4000×5000

B=2×4π

B=8×3.14

B=25.12T

Hence the magnetic field inside the solenoid is  B=25.12T.

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