An iron rod of volume 10-3m3 and relative permeability 1000 is placed as core in a solenoid with 10 turns/cm . If a current of 0.5A is passed through the solenoid, then the magentic moment of rod will be:
Option: 1 500 \times 10^{2}Am^{2}
Option: 2 0.5 \times 10^{2}Am^{2}
Option: 3 50 \times 10^{2}Am^{2}
Option: 4 5 \times 10^{2}Am^{2}

Answers (1)

Correct Answer: c) 50 \times 10^{2}Am^{2}

\begin{aligned} &\text { Here, } n=10 \text { turns } / \mathrm{cm}=1000 \text { turns } / \mathrm{m}\\ &V=10^{-3} m^{3}, \mu_{r}=1000, i=0 \cdot 5 A\\ &\text { We know that } B=\mu_{0}(H+I)\\ &, I=\frac{B}{\mu_{0}}-H=\frac{\mu H}{\mu_{0}}-H\\ &=\mu_{r} H-H=\left(\mu_{r}-1\right) H \end{aligned}

\\ \text{For a solenoid of n turns per unit length carrying current i,} H=n i . \\ \quad \therefore I=\left(\mu_{r}-1\right) n i$ \\ $I=(1000-1) \times 1000 \times 0 \cdot 5$ \\ $I= 5 \times 10^{5} \mathrm{Am}^{-1}$ \\ As magnetic moment, $M=I \times V$ \\ $\therefore M= 5 \times 10^{5} \times 10^{-3}=500 \mathrm{Am}^{2}$

5 \times 10^2 A-m^2

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