The magnetic field strength of copper is $10^6 A/m$ and magnetic susceptibility is $-0.8 × 10^{-3}$. -

Given:-

Magnetic Field Strength, $H = 10^6 A/m$

Magnetic Susceptibility, $χ = -0.8 × 10^{-3}$

To Find:-

Magnetic Flux Density (B)

Magnetization (M)

Solution:-

We know that,

$μ_r = χ + 1$

$μ_r = -0.8 × 10^-3 + 1$

$∴ μ_r = 0.9998$

Now,

$$B = μ_0 μ_r H$$ …