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,

$$B = μ_0 μ_r H$$

$$∴ B =4π × 10^{-7} × 0.9998 × 10^6$$

$$∴ B = 1.2564 Wb/m^2$$

And,

$$M = χH$$

$$∴ M = -0.8 × 10^{-3} × 10^6$$

$$∴ M = -0.8 × 10^3$$