An iron rod is placed parallel to a magnetic | vedclass

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(C) Given:Magnetic field intensity,$H = 2000 \ A \ m^{-1} = 2 	imes 10^3 \ A \ m^{-1}$.Magnetic flux,$\phi = 6 	imes 10^{-4} \ Wb$.Cross-sectional a

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$10^{-3}$

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(C) Given:Magnetic field intensity,$H = 2000 \ A \ m^{-1} = 2 	imes 10^3 \ A \ m^{-1}$.Magnetic flux,$\phi = 6 	imes 10^{-4} \ Wb$.Cross-sectional area,$A = 3 \ cm^2 = 3 	imes 10^{-4} \ m^2$.We know that magnetic flux $\phi = B \cdot A$,where $B$ is the magnetic flux density.So,$B = \frac{\phi}{A} = \frac{6 	imes 10^{-4} \ Wb}{3 	imes 10^{-4} \ m^2} = 2 \ T$.The magnetic permeability $\mu$ is defined as $\mu = \frac{B}{H}$.Substituting the values: $\mu = \frac{2 \ T}{2 	imes 10^3 \ A \ m^{-1}} = 10^{-3} \ Wb \ A^{-1} \ m^{-1}$.

An iron rod is placed parallel to a magnetic field of intensity $2000 \ A \ m^{-1}$. The magnetic flux through the rod is $6 \times 10^{-4} \ Wb$ and its cross-sectional area is $3 \ cm^2$. The magnetic permeability of the rod in $Wb \ A^{-1} \ m^{-1}$ is:

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