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(D) The magnetic field inside the solenoid is $B_0 = \mu_0 n i$.When an iron rod is placed inside,the magnetic field becomes $B = \mu_r B_0 = \mu_r \m

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$5 	imes 10^2$

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(D) The magnetic field inside the solenoid is $B_0 = \mu_0 n i$.When an iron rod is placed inside,the magnetic field becomes $B = \mu_r B_0 = \mu_r \mu_0 n i$.The magnetization $M_z$ is given by $M_z = \frac{B - \mu_0 H}{\mu_0} = (\mu_r - 1) H$.Since $\mu_r = 1000 \gg 1$,we can approximate $M_z \approx \mu_r H = \mu_r n i$.The magnetic moment $m$ is given by $m = M_z 	imes V$,where $V$ is the volume of the rod.Given: $V = 10^{-3} \ m^3$,$\mu_r = 1000$,$n = 10 \ turns/cm = 1000 \ turns/m$,$i = 0.5 \ A$.$m = (\mu_r n i) V = 1000 	imes 1000 	imes 0.5 	imes 10^{-3}$.$m = 10^6 	imes 0.5 	imes 10^{-3} = 0.5 	imes 10^3 = 500 \ Am^2$.$m = 5 	imes 10^2 \ Am^2$.

An iron rod of volume $10^{-3} \ m^3$ and relative permeability $1000$ is placed as a core in a solenoid with $10 \ turns/cm$. If a current of $0.5 \ A$ is passed through the solenoid,then the magnetic moment of the rod will be $........... \ Am^2$.

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