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(C) The magnetic flux $\phi$ is given by $\phi = B \cdot A$,where $B$ is the magnetic field and $A$ is the cross-sectional area.$B = \frac{\phi}{A} =

Vedclass · Accepted Answer

$\mu = 5 \pi 	imes 10^{-4}, \chi = 1249$

Vedclass · Answer

(C) The magnetic flux $\phi$ is given by $\phi = B \cdot A$,where $B$ is the magnetic field and $A$ is the cross-sectional area.$B = \frac{\phi}{A} = \frac{2.4 \pi 	imes 10^{-5} \ Wb}{2 	imes 10^{-5} \ m^2} = 1.2 \pi \ T$.Now,the relation between magnetic field $B$ and magnetising field $H$ is $B = \mu H$.$\mu = \frac{B}{H} = \frac{1.2 \pi \ T}{2400 \ A/m} = 5 \pi 	imes 10^{-4} \ T \cdot m/A$.The relative permeability $\mu_r$ is given by $\mu_r = \frac{\mu}{\mu_0}$.$\mu_r = \frac{5 \pi 	imes 10^{-4}}{4 \pi 	imes 10^{-7}} = 1250$.Since $\mu_r = 1 + \chi$,the magnetic susceptibility $\chi$ is $\chi = \mu_r - 1 = 1250 - 1 = 1249$.

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