A material of 0.25 cm^2 cross-sectional area | vedclass

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(C) The magnetic flux $\phi$ is given by the formula $\phi = B \cdot A$. Since $B = \mu H$ and $\mu = \mu_0(1 + \chi_m)$,we have $\phi = \mu_0(1 + \ch

Vedclass · Accepted Answer

$9.87 	imes 10^{-6} \ Wb$

Vedclass · Answer

(C) The magnetic flux $\phi$ is given by the formula $\phi = B \cdot A$. Since $B = \mu H$ and $\mu = \mu_0(1 + \chi_m)$,we have $\phi = \mu_0(1 + \chi_m) H A$. Given: Area $A = 0.25 \ cm^2 = 0.25 	imes 10^{-4} \ m^2$. Magnetic field strength $H = 1000 \ Am^{-1}$. Susceptibility $\chi_m = 313$. Permeability of free space $\mu_0 = 4 \pi 	imes 10^{-7} \ Hm^{-1}$. Substituting the values: $\phi = (4 \pi 	imes 10^{-7}) 	imes (1 + 313) 	imes 1000 	imes (0.25 	imes 10^{-4})$ $\phi = (4 \pi 	imes 10^{-7}) 	imes 314 	imes 10^3 	imes 0.25 	imes 10^{-4}$ $\phi = 4 \pi 	imes 314 	imes 0.25 	imes 10^{-8}$ $\phi = 314 \pi 	imes 10^{-8} \approx 986.45 	imes 10^{-8} \approx 9.87 	imes 10^{-6} \ Wb$.

$A$ material of $0.25 \ cm^2$ cross-sectional area is placed in a magnetic field of strength $(H) 1000 \ Am^{-1}$. Then,the magnetic flux produced is (Susceptibility of material is $313$) (Permeability of free space,$\mu_0 = 4 \pi \times 10^{-7} \ Hm^{-1}$)

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