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

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$10^{-3}$

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

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

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