Question #191601

A toroidal sample of magnetic material of susceptibility x = 2 × 10^-2 is wound with 100 turns of wire carrying a current of 2 A. The toroid is 0.10 m long.

(a) Find the solenoidal current density.(2)

(b) Determine the magnetic field intensity H produced by the current.(2)

(c) Calculate μ, the magnetic permeability of the material.(2)

(d) Calculate the induced magnetization M in the material.(2)

(e) Calculate the magnetic field B resulting from the current and the magnetization of the

material


1
Expert's answer
2021-05-21T17:06:33-0400

Answer

A.

Current density

j=IA=2π(0.1)3=63.69A/m2j=\frac{I}{A}=\frac{2}{\pi (0.1) ^3}=63.69A/m^2


B.

Magnetic field

B=(1+x)NI2πRB=\frac{(1+x)NI}{2\pi R}



B=(1+0.2)(100)(2)2π(0.1)=324.67TB=\frac{(1+0.2)(100)(2)}{2\pi (0.1)}=324.67T


And

H=B1+x=324.671+0.02=318.3TH=\frac{B}{1+x}=\frac{324.67}{1+0.02}=318.3T



C.

Magnetic permeability

μ=1+x=1+0.02=1.02\mu=1+x=1+0.02=1.02


D.

M=HxM=318.4×0.02=6.5M=Hx \\M=318.4\times0.02=6.5


E.

Magnetic field we have calculated previously

B=324.67TB=324.67T




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Comments

Rk
27.05.21, 13:12

Great answer

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