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  • All straight wires are very long. Both AB and CD are arcs of the same circle, both sub-tending right angles at the centre O. Then the magnetic field at O is <img alt="" src="https://cdn.e

All straight wires are very long. Both AB and CD are arcs of the same circle, both sub-tending right angles at the centre O. Then the magnetic field at O is

Option: 1

\frac{\mu_0i}{4 \pi R}
 

 


Option: 2

\frac{3\mu_0i}{4 \pi R}


Option: 3

\frac{\mu_0i}{2 \pi R}

 


Option: 4

\frac{\mu_0i}{2 \pi R}\left ( \pi +1 \right )


Answers (1)

best_answer

Answer (3)

Field due to AA'=\frac{\mu _{0}i}{4\pi R}

= Field due to BB'

Field due to CC' - field due to DD'=0

Field due to BA=\frac{\mu _{0}i}{8R}

Field due to CD=\frac{\mu _{0}i}{8R}

\therefore Net field at O=\frac{\mu _{0}i}{2 \pi R}

Posted by

Suraj Bhandari

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