An iron rod of cross-sectional area 2 times 1 | vedclass

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Question

(B) The magnetic flux density $B$ is given by the ratio of magnetic flux $\phi$ to the cross-sectional area $A$:$B = \frac{\phi}{A} = \frac{6.28 	ime

Vedclass · Accepted Answer

$1.25 	imes 10^4$

Vedclass · Answer

(B) The magnetic flux density $B$ is given by the ratio of magnetic flux $\phi$ to the cross-sectional area $A$:$B = \frac{\phi}{A} = \frac{6.28 	imes 10^{-4} \ Wb}{2 	imes 10^{-5} \ m^2} = 31.4 \ T$The relationship between magnetic flux density $B$,permeability $\mu$,and magnetizing field intensity $H$ is $B = \mu H$.Therefore,the permeability $\mu$ is:$\mu = \frac{B}{H} = \frac{31.4 \ T}{2000 \ A/m} = 0.0157 \ T \cdot m/A$However,calculating the relative permeability $\mu_r$ using $\mu = \mu_0 \mu_r$:$\mu_r = \frac{\mu}{\mu_0} = \frac{0.0157}{4\pi 	imes 10^{-7}} \approx 1.25 	imes 10^4$Thus,the magnetic permeability $\mu$ is approximately $1.25 	imes 10^4 \ \mu_0$.

An iron rod of cross-sectional area $2 \times 10^{-5} \ m^2$ is subjected to a magnetizing field intensity of $2000 \ A/m$. If the magnetic flux produced in the rod is $6.28 \times 10^{-4} \ Wb$,what is the magnetic permeability of the rod?

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