Question Number 40946 by Necxx last updated on 29/Jul/18
$${Each}\:{of}\:{two}\:{long}\:{straight}\:{wires}\:\mathrm{4}{cm} \\ $$$${apart}\:{carry}\:{equal}\:{electric}\:{currents} \\ $$$${and}\:{experience}\:{a}\:{force}\:{of}\:\mathrm{2}×\mathrm{10}^{−\mathrm{4}} {N}/{m}. \\ $$$${What}\:{is}\:{the}\:{magnitude}\:{of}\:{the} \\ $$$${electric}\:{current}\:{in}\:{each}? \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 30/Jul/18
$$\frac{{F}}{{l}}=\frac{\mu_{\mathrm{0}} {i}_{\mathrm{1}} {i}_{\mathrm{2}} }{\mathrm{2}\Pi{a}}\:\:\:\frac{{F}}{{l}}={force}\:{per}\:{uni}\:{length} \\ $$$$\mathrm{2}×\mathrm{10}^{−\mathrm{4}} =\frac{\mu_{\mathrm{0}} {i}^{\mathrm{2}} }{\mathrm{2}\Pi×.\mathrm{04}}\:\:\:{cslculate}\:… \\ $$