An iron rod is placed parallel to a magnetic | vedclass

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(B) Given:Magnetic field intensity, $H = 1000 	ext{ A/m}$Magnetic flux, $\phi = 3 	imes 10^{-4} 	ext{ Wb}$Cross-sectional area, $A = 1.5 	ext{ cm}

Vedclass · Accepted Answer

$2 	imes 10^{-3}$

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(B) Given:Magnetic field intensity, $H = 1000 	ext{ A/m}$Magnetic flux, $\phi = 3 	imes 10^{-4} 	ext{ Wb}$Cross-sectional area, $A = 1.5 	ext{ cm}^2 = 1.5 	imes 10^{-4} 	ext{ m}^2$First, calculate the magnetic flux density $(B)$:$B = \frac{\phi}{A} = \frac{3 	imes 10^{-4} 	ext{ Wb}}{1.5 	imes 10^{-4} 	ext{ m}^2} = 2 	ext{ T (or Wb/m}^2)$The relationship between magnetic flux density $(B)$, magnetic permeability $(\mu)$, and magnetic field intensity $(H)$ is given by:$B = \mu H$Therefore, the magnetic permeability $(\mu)$ is:$\mu = \frac{B}{H} = \frac{2 	ext{ Wb/m}^2}{1000 	ext{ A/m}} = 2 	imes 10^{-3} 	ext{ Wb/(A} \cdot 	ext{m)}$Thus, the correct option is $B$.

An iron rod is placed parallel to a magnetic field intensity of $1000 \text{ A/m}$. The magnetic flux through the rod is $3 \times 10^{-4} \text{ Wb}$ and its cross-sectional area is $1.5 \text{ cm}^2$. The magnetic permeability of the rod in $\text{Wb/(A} \cdot \text{m)}$ is:

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