Three long straight wires are connected parallel to each other across a battery of negligible internal resistance. The ratio of their resistances are $3 : 4 : 5$. What is the ratio of distances of middle wire from the others if the net force experienced by it is zero
Diffcult
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(c) The wires are in parallel and ratio of their resistances are $3 : 4 : 5$, Hence currents in wires are in the ratio $\frac{1}{3}:\frac{1}{4}:\frac{1}{5}$
${i_1} = \frac{k}{3}$
${i_{_2}} = \frac{k}{4}$
${i_3} = \frac{k}{5}$
Force between top and middle wire ${F_1} = \frac{{{\mu _0}}}{{4\pi }}.\frac{{2{i_1}{i_2}}}{{{r_1}}}$
$ = \frac{{{\mu _0}}}{{4\pi }} \times \frac{{2\left( {\frac{1}{3}} \right)\,\left( {\frac{1}{4}} \right)\,{k^2}}}{{{r_1}}}$.
${i_1} = \frac{k}{3}$
${i_{_2}} = \frac{k}{4}$
${i_3} = \frac{k}{5}$
Force between top and middle wire ${F_1} = \frac{{{\mu _0}}}{{4\pi }}.\frac{{2{i_1}{i_2}}}{{{r_1}}}$
$ = \frac{{{\mu _0}}}{{4\pi }} \times \frac{{2\left( {\frac{1}{3}} \right)\,\left( {\frac{1}{4}} \right)\,{k^2}}}{{{r_1}}}$.
Force between bottom and middle wire ${F_2} = \frac{{{\mu _0}}}{{4\pi }}\frac{{\left( {\frac{1}{4}} \right)\;\left( {\frac{1}{5}} \right){k^2}}}{{{r_2}}}$.
As the forces are equal and opposite so ${F_1} = {F_2} \Rightarrow \frac{{{r_1}}}{{{r_2}}} = \frac{5}{3}$.
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