A magnetizing field of 100 ,A/m produces a ma | vedclass

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Question

$(A)$ The magnetic induction $B$ is given by the ratio of magnetic flux $\phi$ to the cross-sectional area $A$: $B = \frac{\phi}{A}$.Given: $\phi = 2.

Vedclass · Accepted Answer

$8 	imes 10^{-4}$

Vedclass · Answer

$(A)$ The magnetic induction $B$ is given by the ratio of magnetic flux $\phi$ to the cross-sectional area $A$: $B = \frac{\phi}{A}$.Given: $\phi = 2.4 	imes 10^{-5} \,Wb$ and $A = 0.3 \,cm^2 = 0.3 	imes 10^{-4} \,m^2$.Calculating $B$: $B = \frac{2.4 	imes 10^{-5}}{0.3 	imes 10^{-4}} = \frac{2.4}{0.3} 	imes 10^{-1} = 8 	imes 0.1 = 0.8 \,T$ (or $Wb/m^2$).The magnetic permeability $\mu$ is defined as the ratio of magnetic induction $B$ to the magnetizing field intensity $H$: $\mu = \frac{B}{H}$.Given: $H = 100 \,A/m$.Calculating $\mu$: $\mu = \frac{0.8}{100} = 8 	imes 10^{-3} \,T \cdot m/A$. Wait, re-evaluating the calculation: $B = 0.8 \,T$. $\mu = \frac{0.8}{100} = 0.008 = 8 	imes 10^{-3}$.Reviewing the provided options, there is a discrepancy. If $H = 1000 \,A/m$, then $\mu = 8 	imes 10^{-4}$. Assuming the intended value was $1000 \,A/m$, the correct option is $A$.

$A$ magnetizing field of $100 \,A/m$ produces a magnetic flux of $2.4 \times 10^{-5} \,Wb$ in an iron bar of cross-sectional area $0.3 \,cm^2$. The magnetic permeability of the iron bar in the $SI$ unit is

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