The magnetic field at the centre of a circula | vedclass

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Question

(a) Case $1$ : ${B_A} = \frac{{{\mu _0}}}{{4\pi }}.\frac{i}{r} \otimes $ 

${B_B} = \frac{{{\mu _0}}}{{4\pi }}.\frac{{\pi \,i}}{r}\odot$

Vedclass · Accepted Answer

$\left( { - \frac{\pi }{2}} \right):\left( {\frac{\pi }{2}} \right):\left( {\frac{{3\pi }}{4} - \frac{1}{2}} \right)$

Vedclass · Answer

(a) Case $1$ : ${B_A} = \frac{{{\mu _0}}}{{4\pi }}.\frac{i}{r} \otimes $ 

${B_B} = \frac{{{\mu _0}}}{{4\pi }}.\frac{{\pi \,i}}{r}\odot$ 

${B_C} = \frac{{{\mu _0}}}{{4\pi }}.\frac{i}{r}\odot$ 

So net magnetic field at the centre of case $1$ 

${B_1} = {B_B} - {B_C} - {B_A} \Rightarrow {B_1} = \frac{{{\mu _0}}}{{4\pi }}.\frac{{\pi i}}{r}\odot$..... $(i)$

Case $2$ : As we discussed before magnetic field at the centre $O$ in this case 

${B_2} = \frac{{{\mu _0}}}{{4\pi }}.\frac{{\pi i}}{r} \otimes $ ..... $(ii)$

Case $3$ : ${B_A} = 0$ ${B_B} = \frac{{{\mu _0}}}{{4\pi }}.\frac{{(2\pi - \pi /2)i}}{r} \otimes $ 

${B_C} = \frac{{{\mu _0}}}{{4\pi }}.\frac{i}{r}\odot$

$ = \frac{{{\mu _0}}}{{4\pi }}.\frac{{3\pi i}}{{2r}} \otimes $ 

So net magnetic field at the centre of case $3$ ${B_3} = \frac{{{\mu _0}}}{{4\pi }}.\frac{i}{r}\left( {\frac{{3\pi }}{2} - 1} \right) \otimes $..... $(iii)$

From equation $(i)$, $(ii)$ and $(iii)$ 

${B_1}:{B_2}:{B_3} = \pi \odot$ : $\pi \otimes $ $\left( {\frac{{3\pi }}{2} - 1} \right)\, \otimes = - \frac{\pi }{2}:\frac{\pi }{2}:\left( {\frac{{3\pi }}{4} - \frac{1}{2}} \right)$

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