Physics, asked by saiharshitham9794, 9 months ago

A paramagnetic material has a magnetic field intensity of 104 A/m. If the susceptibility of the material at room temperature is 3.7 x 10-3, calculate the magnetization and magnetic flux density in the material.

Answers

Answered by SKBhattacharjee54
9

Hey mate here ur answer

Given: χ = –0.4 × 10‒5

H = 5 × 105 A /m

B = ? and M = ?

B = μ0(H + M) = μ0 H(1 + χ)

= 4π × 10‒7 × 5 × 105 [1 – 0.4 × 10‒5] = 4π × 5 × 10‒2 × 0.9996 = 0.62 Wb/m2

M = χH = –0.4 × 10–5 × 5 × 105 = –2.0 A/m.

2. The magnetic field strength in silicon is 1000 A/m. If the magnetic susceptibility is −0.25 × 10−5, calculate the magnetization and flux density in silicon.

sol: Magnetic field strength (H) = 1000 A/m

Magnetic susceptibility (χ) = –0.25 × 10‒5

Magnetization (M) = χH = –0.25 × 10‒5 × 1000 = –0.25 × 10‒2 A/m.

=>Hope it helps u....

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Answered by Tulsi4890
4

Given:

A paramagnetic material with magnetic field intensity (H) = 10⁴ A/m

The susceptibility at room temperature (Χ) = 3.7 X 10⁻³

To Find:

The magnetization (M) and magnetic flux density (B) in the material

Solution:

Paramagnetic materials have at least one unpaired electron and are weakly attracted by an externally applied magnetic field.

We know that magnetic susceptibility is the ratio of magnetization to the magnetic field intensity.

i.e.  X = M / H

or M = X.H

or M = 3.7 X 10⁻³ X 10⁴  

= 37 A/m

Flux Density = B = μo (M + H)

Here μo = 10⁻⁷ X 4π

Substituting,

B = 10⁻⁷ X 4π ( 37 + 10⁴ )

Taking π = 3.14,

B = 10⁻⁷ X 12.56 X ( 37 + 10⁴ )

= 126179.4 X 10⁻⁷

= 1.26 X 10⁻² Wbm⁻²

Hence, the magnetization of the given material is 37A/m and the magnetic flux density is 1.26 X 10⁻² Wbm⁻².

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