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Determination of the magnetization \(M_x\)

Along the axis of symmetry of the magnet the following applies:

$$ { B_x(x) = \frac{1}{2} M_x \mu_0 \left( \frac{\frac{L}{2}-x}{\sqrt{\left(\frac{L}{2} - x\right)^2 + R^2}} + \frac{\frac{L}{2} + x}{\sqrt{\left(\frac{L}{2} + x\right)^2 + R^2}} \right)} \\$$

Task: Obtain a value for the magnetization. Use the slider to adjust the formula to match the measurement.

\(M_x=\)\(\frac{\mathrm{kA}}{\mathrm{m}}\)

Derivation of the formula

\(\mu_0 = 1,2566 \cdot 10^{-6} \frac{\mathrm{N}}{\mathrm{A^2}} ,\\ R=7,5\mathrm{mm}, \\L=100\mathrm{mm}\)

Skizze Magnet

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