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  • Please help! 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:

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}\text{Il}}{3\pi}

  • Option 2)

    \frac{\mu_{0}\text{Il}}{2\pi}

  • Option 3)

    \frac{2\mu_{0}\text{IiL}}{3\pi}

  • Option 4)

    \frac{\mu_{0}\text{IiL}}{2\pi}

 

Answers (1)

best_answer

 

Force between two parallel current carrying conductors -

F=frac{mu }{4pi } frac{2I_{1 I_{2}}}{a} l

frac{F}{l}=frac{mu o}{4pi } frac{2I_{1 I_{2}}}{a}

- wherein

I1 and I2 current carrying two parallel wires 

a-seperation between two wires 

 

 force on F_{1}= \frac{\mu _{0}}{4\pi }\frac{2Iil}{\frac{l}{2}} = \frac{\mu _{0} Ii}{3\pi }

force  f_{2 }= \frac{\mu _{0}}{4\pi }\frac{2IiL}{3\left ( \frac{L}{2} \right )} = \frac{\mu_{0} }{3\pi }

Net force on the loop

f_{1}-f_{2} = \frac{\mu _{0Ii}}{\pi }\left [ 1-\frac{1}{3} \right ]

= \frac{2 \mu _{0}Ii}{3\pi }


Option 1)

\frac{2\mu_{0}\text{Il}}{3\pi}

This option is correct.

Option 2)

\frac{\mu_{0}\text{Il}}{2\pi}

This option is incorrect.

Option 3)

\frac{2\mu_{0}\text{IiL}}{3\pi}

This option is incorrect.

Option 4)

\frac{\mu_{0}\text{IiL}}{2\pi}

This option is incorrect.

Posted by

prateek

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