क्षेत्रफल A और प्रतिरोध R का एक छोटा वृत्ताका | vedclass

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(C) सोलेनोइड का चुंबकीय क्षेत्र $\vec{B} = \mu_0 m I \hat{n}$ है। लूप $xy$-समतल में है,इसलिए इसका क्षेत्रफल सदिश $\vec{A} = A\hat{k}$ है।$(I)$ $\hat{n

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$I \rightarrow Q, II \rightarrow P, III \rightarrow S, IV \rightarrow R$

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(C) सोलेनोइड का चुंबकीय क्षेत्र $\vec{B} = \mu_0 m I \hat{n}$ है। लूप $xy$-समतल में है,इसलिए इसका क्षेत्रफल सदिश $\vec{A} = A\hat{k}$ है।$(I)$ $\hat{n} = \frac{1}{\sqrt{2}}(\sin \omega t \hat{j} + \cos \omega t \hat{k})$। फ्लक्स $\phi = \vec{B} \cdot \vec{A} = \frac{\mu_0 m I A}{\sqrt{2}} \cos \omega t$। प्रेरित $EMF$ $\varepsilon = -\frac{d\phi}{dt} = \frac{\mu_0 m I A \omega}{\sqrt{2}} \sin \omega t$। धारा $i = \frac{\varepsilon}{R}$। चुंबकीय आघूर्ण $\vec{M} = i A \hat{k} = \frac{\mu_0 m I A^2 \omega}{\sqrt{2} R} \sin \omega t \hat{k}$। टॉर्क $\vec{	au} = \vec{M} 	imes \vec{B} = \frac{\mu_0^2 m^2 I^2 A^2 \omega}{2R} \sin^2 \omega t (\hat{k} 	imes \hat{j}) = \alpha \sin^2 \omega t (-\hat{i})$। $t = \frac{\pi}{6\omega}$ पर,$\sin^2(\pi/6) = 1/4$,इसलिए $\vec{	au} = -\frac{\alpha}{4} \hat{i}$ $(Q)$।$(II)$ $\hat{n} = \frac{1}{\sqrt{2}}(\sin \omega t \hat{i} + \cos \omega t \hat{j})$। यहाँ $\vec{B} \cdot \hat{k} = 0$,इसलिए $\phi = 0$,$\varepsilon = 0$,$i = 0$,$\vec{	au} = 0$ $(P)$।$(III)$ $\hat{n} = \frac{1}{\sqrt{2}}(\sin \omega t \hat{i} + \cos \omega t \hat{k})$। फ्लक्स $\phi = \frac{\mu_0 m I A}{\sqrt{2}} \cos \omega t$। $(I)$ की तरह,$i = \frac{\mu_0 m I A \omega}{\sqrt{2} R} \sin \omega t$। $\vec{M} = i A \hat{k}$। $\vec{	au} = \vec{M} 	imes \vec{B} = \alpha \sin^2 \omega t (\hat{k} 	imes \hat{i}) = \alpha \sin^2 \omega t \hat{j}$। $t = \frac{\pi}{6\omega}$ पर,$\vec{	au} = \frac{\alpha}{4} \hat{j}$ $(S)$।$(IV)$ $\hat{n} = \frac{1}{\sqrt{2}}(\cos \omega t \hat{j} + \sin \omega t \hat{k})$। फ्लक्स $\phi = \frac{\mu_0 m I A}{\sqrt{2}} \sin \omega t$। $\varepsilon = -\frac{d\phi}{dt} = -\frac{\mu_0 m I A \omega}{\sqrt{2}} \cos \omega t$। $i = -\frac{\mu_0 m I A \omega}{\sqrt{2} R} \cos \omega t$। $\vec{M} = i A \hat{k}$। $\vec{	au} = \vec{M} 	imes \vec{B} = \alpha \cos^2 \omega t (-\hat{k} 	imes \hat{j}) = \alpha \cos^2 \omega t \hat{i}$। $t = \frac{\pi}{6\omega}$ पर,$\cos^2(\pi/6) = 3/4$,इसलिए $\vec{	au} = \frac{3\alpha}{4} \hat{i}$ $(R)$।

$List-I$	$List-II$
$(I)$ $\frac{1}{\sqrt{2}}(\sin \omega t \hat{j}+\cos \omega t \hat{k})$	$(P)$ $0$
$(II)$ $\frac{1}{\sqrt{2}}(\sin \omega t \hat{i}+\cos \omega t \hat{j})$	$(Q)$ $-\frac{\alpha}{4} \hat{i}$
$(III)$ $\frac{1}{\sqrt{2}}(\sin \omega t \hat{i}+\cos \omega t \hat{k})$	$(R)$ $\frac{3\alpha}{4} \hat{i}$
$(IV)$ $\frac{1}{\sqrt{2}}(\cos \omega t \hat{j}+\sin \omega t \hat{k})$	$(S)$ $\frac{\alpha}{4} \hat{j}$

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