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  • Two wires 1 and 2 carrying currents i1 and i2 respectively are inclined at an angle q as shown in the figure. What is the force on a small element dl of wire 2 at a dista

Two wires 1 and 2 carrying currents i1 and i2 respectively are inclined at an angle q as shown in the figure. What is the force on a small element dl of wire 2 at a distance r from wire 1 due to the magnetic field of wire 1 ?

Option: 1

\mathrm{\frac{\mu_0}{2 \pi r}\left(i_1 i_2 d l \tan \theta\right)}


Option: 2

\mathrm{\frac{\mu_0}{2 \pi r}\left(i_1 i_2 d l \sin \theta\right)}


Option: 3

\mathrm{\frac{\mu_0}{2 \pi r}\left(i_1 i_2 d l \cos \theta\right)}


Option: 4

\mathrm{\frac{\mu_0}{4 \pi r}\left(i_1 i_2 d l \sin \theta\right)}


Answers (1)

best_answer

The component of dl parallel to wire 1 is \mathrm{d l \cos \theta}. Hence the force on the element is
\mathrm{ F=B_1 i_2 d l \cos \theta }

where   \mathrm{ B_1 } =  magnetic field at the element due to current in wire which is given by
\mathrm{ B_1 =\frac{\mu_0 i_1}{2 \pi r} \\ }
\mathrm{ \therefore F =\frac{\mu_0}{2 \pi r}\left(i_1 i_2 d l \cos \theta\right) }
 

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Sayak

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