The relation between total magnetic field (B) | vedclass

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(B) The net magnetic field $(B)$ inside a material is the sum of the magnetic field due to external current $(B_0)$ and the magnetic field due to the

Vedclass · Accepted Answer

$\frac{B}{H}=\mu_0(1+\chi)$

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(B) The net magnetic field $(B)$ inside a material is the sum of the magnetic field due to external current $(B_0)$ and the magnetic field due to the magnetization of the material $(B_m)$.$B = B_0 + B_m$We know that $B_0 = \mu_0 H$, where $H$ is the magnetic intensity.The magnetization $(M)$ of the material is related to the magnetic intensity by $M = \chi H$, where $\chi$ is the magnetic susceptibility.The magnetic field due to magnetization is $B_m = \mu_0 M = \mu_0 \chi H$.Substituting these into the expression for $B$:$B = \mu_0 H + \mu_0 \chi H$$B = \mu_0 H(1 + \chi)$Dividing both sides by $H$, we get:$\frac{B}{H} = \mu_0(1 + \chi)$

The relation between total magnetic field $(B)$, magnetic intensity $(H)$, permeability of free space $(\mu_0)$, and susceptibility $(\chi)$ is:

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