Question

plzzzz answer it...
I don't need only option I need the explanation...
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Answer 1

Answer:

hope the answer will help you

Answer 2

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❏ Question:-

Find the Magnetic Intensity at the point O due to the whole loop

Fig :-

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❏ Solution:-

$\blacksquare\:\:\:\underline\red{\text{For AB , M.I. at point O is }}$

$\longrightarrow \theta_1=0\degree\:\:and\:\:\theta_2=45\degree$

$\therefore \:B_{AB} =\dfrac{\mu_oi}{4\pi b}(\sin0\degree+\sin45\degree)$

$\implies B_{AB} =\dfrac{\mu_oi}{4\pi b}(0+\dfrac{1}{\sqrt{2}})$

$\implies B_{AB} =\dfrac{\mu_oi}{4\sqrt{2}\pi b}$ ( downward )

$\blacksquare\:\:\:\underline\red{\text{For BC , M.I. at point O is }}$

$\longrightarrow \theta_2=0\degree\:\:and\:\:\theta_1=45\degree$

So , similarly

$\implies B_{BC} =\dfrac{\mu_oi}{4\sqrt{2}\pi b}$ ( downward )

$\blacksquare\:\:\:\underline\red{\text{For CD and AF , M.I. at point O is }}$

$\longrightarrow \theta_1=0\:\:and\:\:\theta_2=0\degree$

$\therefore \:\:B_{CD}=B_{AF} =\dfrac{\mu_oi}{4\pi b}(\sin0\degree+\sin0\degree)$

$\implies B_{CD}=B_{AF} =\dfrac{\mu_oi}{4\pi b}\times 0$

$\implies B_{CD}=B_{AF} = 0$

$\blacksquare\:\:\:\underline\red{\text{For the circle DF , M.I. at point O is }}$

$\longrightarrow \theta_1=270\degree=\dfrac{3\pi}{2}$

$\therefore\:\: B_{DF} =\dfrac{\mu_oi}{4\cancel{\pi}a}\times\dfrac{3\cancel{\pi}}{2}$

$\implies B_{DF} =\dfrac{3\mu_oi}{8a}$ ( downward )

$\blacksquare\:\:\:\underline\red{\text{So the net , M.I. at point O is }}$

$\therefore\:\: B=B_{AB}+B_{BC}+B_{DF}$

$\implies B=\dfrac{\mu_oi}{4\sqrt{2}\pi b}+\dfrac{\mu_oi}{4\sqrt{2}\pi b}+\dfrac{3\mu_oi}{8a}$

$\implies\boxed{ B=\dfrac{\mu_oi}{4\pi}\Big(\dfrac{3\pi}{2a}+\dfrac{\sqrt{2}}{b}\Big)}$ (downward)

❏ Used Theory

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