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  • A square loop ABCD carrying a current i, is placed near and coplanar with a long straight conductor XY carrying a current I, the net force on the loop will be : <img alt="" src="http

A square loop ABCD carrying a current i, is placed near and coplanar with a long straight conductor XY carrying a current I, the net force on the loop will be :

Option: 1

\frac{2 \mu_0 \mathrm{Ii}}{3 \pi}


Option: 2

\frac{\mu_0 \mathrm{Ii}}{2 \pi}


Option: 3

\frac{2 \mu_0 \mathrm{IiL}}{3 \pi}


Option: 4

\frac{\mu_0 \mathrm{IiL}}{2 \pi}


Answers (1)

best_answer

The direction of current in conductor 

XY and AB is same
\therefore \mathrm{F}_{\mathrm{AB}}=\mathrm{i}
B (attractive)
\mathrm{\mathrm{F}_{A B}=\mathrm{i}(\mathrm{L}) \cdot \frac{\mu_0 \mathrm{I}}{2 \pi\left(\frac{L}{2}\right)}(\leftarrow)=\frac{\mu_0 \mathrm{iI}}{\pi}(\leftarrow)}

\mathrm{ \mathrm{F}_{\mathrm{BC}} \text { opposite to } \mathrm{F}_{\mathrm{AD}} \\ }

\mathrm{ \mathrm{F}_{\mathrm{BC}}(\uparrow) \text { and } \mathrm{F}_{\mathrm{AD}}(\downarrow) \\ }

\mathrm{ \Rightarrow \text { cancels each other } }

\mathrm{ \mathrm{F}_{\mathrm{CD}}=\mathrm{i} \\ }

(B repulsive)

\mathrm{ \mathrm{F}_{\mathrm{CD}}=\mathrm{i}(\mathrm{L}) \frac{\mu_0 \mathrm{I}}{2 \pi\left(\frac{\mathrm{3L}}{2}\right)}(\rightarrow)=\frac{\mu_0 \mathrm{iI}}{3 \pi}(\rightarrow) }

Therefore the net force on the loop
\mathrm{ \mathrm{F}_{\mathrm{net}}=\mathrm{F}_{\mathrm{AB}}+\mathrm{F}_{\mathrm{BC}}+\mathrm{F}_{\mathrm{CD}}+\mathrm{F}_{\mathrm{AD}} \\ }

\mathrm{ \Rightarrow \mathrm{F}_{\mathrm{nct}}=\frac{\mu_{\mathrm{o}} \mathrm{iI}}{\pi}-\frac{\mu_0 \mathrm{iI}}{3 \pi}=\frac{2 \mu_0 \mathrm{iI}}{3 \pi} }
 

Posted by

sudhir kumar

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