An iron rod of volume 10^{-4} , m^3 and relat | vedclass

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(D) The magnetic intensity $H$ inside the solenoid is given by $H = ni$.Given $n = 5 \, 	ext{turns/cm} = 500 \, 	ext{turns/m}$ and $i = 0.5 \, A$.So

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$25$

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(D) The magnetic intensity $H$ inside the solenoid is given by $H = ni$.Given $n = 5 \, 	ext{turns/cm} = 500 \, 	ext{turns/m}$ and $i = 0.5 \, A$.So, $H = 500 	imes 0.5 = 250 \, A/m$.The intensity of magnetization $I$ is given by $I = (\mu_r - 1)H$.Given $\mu_r = 1000$, so $I = (1000 - 1) 	imes 250 = 999 	imes 250 = 249750 \, A/m \approx 2.5 	imes 10^5 \, A/m$.The magnetic moment $M$ is given by $M = I 	imes V$, where $V = 10^{-4} \, m^3$.$M = 2.5 	imes 10^5 	imes 10^{-4} = 25 \, A m^2$.

An iron rod of volume $10^{-4} \, m^3$ and relative permeability $1000$ is placed inside a long solenoid wound with $5 \, \text{turns/cm}$. If a current of $0.5 \, A$ is passed through the solenoid, then the magnetic moment of the rod is .... $A m^2$.

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