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  • An arrangement of three parallel straight wires placed perpendicular to plane of paper carrying same current 'I' along the same direction is shown in fig. Magnitude of force per unit length

An arrangement of three parallel straight wires placed perpendicular to plane of paper carrying same current 'I' along the same direction is shown in fig. Magnitude of force per unit length on the middle wire 'B' is given by

Option: 1

\frac{2 \mu_0 \mathrm{i}^2}{\pi \mathrm{d}}


Option: 2

\frac{\sqrt{2} \mu_0 \mathrm{i}^2}{\pi \mathrm{d}}


Option: 3

\frac{\mu_0 \mathrm{i}^2}{\sqrt{2} \pi \mathrm{d}}


Option: 4

\frac{\mu_0 \mathrm{i}^2}{2 \pi \mathrm{d}}


Answers (1)

best_answer

Force per unit length between two parallel current carrying conductors,
\mathrm{F}=\frac{\mu_0 \mathrm{i}_1 \mathrm{i}_2}{2 \pi \mathrm{d}}
Since same current flowing through both the wires

\mathrm{ \mathrm{i}_{\mathrm{i}}=\mathrm{i}_2=\mathrm{i} \\ }

\mathrm{ \text { so } \mathrm{F}_1=\frac{\mu_0 \mathrm{i}^2}{2 \pi \mathrm{d}}=\mathrm{F}_2 }

\mathrm{ \therefore }  Magnitude of force per unit length on the middle wire 'B'
\mathrm{ \mathrm{F}_{\text {net }}=\sqrt{\mathrm{F}_1^2+\mathrm{F}_2^2}=\frac{\mu_0 \mathrm{i}^2}{\sqrt{2} \pi \mathrm{d}} }

Posted by

SANGALDEEP SINGH

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