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Question

(C) The magnetic permeability $\mu$ is defined as the ratio of magnetic induction $B$ to the magnetic field intensity $H$,given by $\mu = \frac{B}{H}$

Vedclass · Accepted Answer

$10^{-3}$

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(C) The magnetic permeability $\mu$ is defined as the ratio of magnetic induction $B$ to the magnetic field intensity $H$,given by $\mu = \frac{B}{H}$.Since magnetic flux $\phi = B \cdot A$,we have $B = \frac{\phi}{A}$.Substituting this into the permeability formula: $\mu = \frac{\phi}{A \cdot H}$.Given values:$\phi = 6 	imes 10^{-4} 	ext{ Wb}$$A = 3 	ext{ cm}^2 = 3 	imes 10^{-4} 	ext{ m}^2$$H = 2000 	ext{ A/m} = 2 	imes 10^3 	ext{ A/m}$Calculating $\mu$:$\mu = \frac{6 	imes 10^{-4}}{(3 	imes 10^{-4}) 	imes (2 	imes 10^3)} = \frac{6 	imes 10^{-4}}{6 	imes 10^{-1}} = 10^{-3} 	ext{ Wb/(A} \cdot 	ext{m)}$.

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

Similar Questions

An iron rod of cross-sectional area $0.2 \, cm^2$ is subjected to a magnetizing field intensity of $1600 \, A \cdot m^{-1}$. If the magnetic flux produced in the rod is $2.4 \times 10^{-5} \, Wb$, what is the magnetic susceptibility of the material of the rod?

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