The relation between total magnetic field (B) | vedclass

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(A) The total magnetic field $(B)$ inside a material is given by the sum of the magnetic field due to external current $(B_{0} = \mu_{0}H)$ and the ma

Vedclass · Accepted Answer

$\frac{B}{H} = \mu_{0}(1+\chi)$

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(A) The total magnetic field $(B)$ inside a material is given by the sum of the magnetic field due to external current $(B_{0} = \mu_{0}H)$ and the magnetic field due to magnetization $(B_{m} = \mu_{0}M)$.Thus,$B = \mu_{0}(H + M)$.We know that magnetization $(M)$ is related to magnetic intensity $(H)$ by the susceptibility $(\chi)$ as $M = \chi H$.Substituting this into the equation for $B$:$B = \mu_{0}(H + \chi H) = \mu_{0}H(1 + \chi)$.Rearranging the terms to find the ratio $\frac{B}{H}$:$\frac{B}{H} = \mu_{0}(1 + \chi)$.

The relation between total magnetic field $(B)$,magnetic intensity $(H)$,permeability of free space $(\mu_{0})$,and magnetic susceptibility $(\chi)$ is:

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