A long solenoid with 1000 , text{turns/m} has | vedclass

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(D) The magnetic moment $M$ of a core material in a solenoid is given by $M = I_{m} V$, where $I_{m}$ is the intensity of magnetization and $V$ is the

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$250$

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(D) The magnetic moment $M$ of a core material in a solenoid is given by $M = I_{m} V$, where $I_{m}$ is the intensity of magnetization and $V$ is the volume.$I_{m} = \chi H$, where $\chi$ is the magnetic susceptibility and $H$ is the magnetic field intensity.Since $\chi = \mu_{r} - 1$, we have $M = (\mu_{r} - 1) H V$.For a long solenoid, $H = nI$, which remains constant.Thus, $M \propto (\mu_{r} - 1)$.The fractional change in magnetic moment is $\frac{\Delta M}{M} = \frac{(\mu_{r2} - 1) - (\mu_{r1} - 1)}{\mu_{r1} - 1} = \frac{\mu_{r2} - \mu_{r1}}{\mu_{r1} - 1}$.Given $\mu_{r1} = 500$ and $\mu_{r2} = 750$, we get $\frac{\Delta M}{M} = \frac{750 - 500}{500 - 1} = \frac{250}{499}$.Comparing this with $\frac{x}{499}$, we find $x = 250$.

$(i)$	$(ii)$	$(iii)$
$(A). \frac{\mu_0 i}{2r} \odot$	$(A). \frac{\mu_0}{2\pi} \frac{i}{r}(\pi - 2)$	$(A). \frac{\mu_0}{2r} \frac{2i}{r}(\pi + 1) \otimes$
$(B). \frac{\mu_0 i}{2r} \otimes$	$(B). \frac{\mu_0 i}{4\pi} \frac{i}{r}(\pi + 2) \otimes$	$(B). \frac{\mu_0 i}{4r} \frac{2i}{r}(\pi - 1) \otimes$
$(C). \frac{3\mu_0 i}{8r} \otimes$	$(C). \frac{\mu_0 i}{4r} \otimes$	$(C). \text{Zero}$
$(D). \frac{3\mu_0 i}{8r} \odot$	$(D). \frac{\mu_0 i}{4r} \odot$	$(D). \text{Infinite}$

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