Question

P and Q are long straight conductors 'r' distance apart . N is a point in the plane of wires r/4 distance away from P carrying current I . What is the magnitude and direction of current in the wire Q,so that net magnetic field at N is 0

Answer 1

$$\begin{gathered} \mathrm{Magnetic}\mathrm{field}\mathrm{produced}\mathrm{by}\mathrm{conductor}\mathrm{P}\mathrm{at}\mathrm{point}\mathrm{N}. \\ {\mathrm{B}}_{\mathrm{P}\mathrm{at}\mathrm{N}}=\frac{{\mathrm{\mu }}_0\mathrm{I}}{2\mathrm{\pi }{\displaystyle \frac{\mathrm{r}}{4}}}=2\frac{{\mathrm{\mu }}_0\mathrm{I}}{\mathrm{\pi r}} \\ \mathrm{The}\mathrm{magnitude}\mathrm{of}\mathrm{the}\mathrm{magnetic}\mathrm{field}\mathrm{at}\mathrm{N}\mathrm{produced}\mathrm{by}\mathrm{the}\mathrm{conductor}\mathrm{Q}\mathrm{should}\mathrm{be}\mathrm{exactly}\mathrm{equal}\mathrm{to} \\ {\mathrm{B}}_{\mathrm{P}\mathrm{at}\mathrm{N}}\mathrm{but}\mathrm{in}\mathrm{opposite}\mathrm{direction}\mathrm{it}\mathrm{negate}\mathrm{its}\mathrm{effect}. \\ \mathrm{Therefore}, \\ {\mathrm{B}}_{\mathrm{Q}\mathrm{at}\mathrm{N}}=\frac{{\mathrm{\mu }}_0\mathrm{I}\text{'}}{2\mathrm{\pi }{\displaystyle \frac{3\mathrm{r}}{4}}}=\frac{2}{3}\frac{{\mathrm{\mu }}_0\mathrm{I}\text{'}}{\mathrm{\pi r}} \\ \mathrm{or}, \\ \frac{2}{3}\frac{{\mathrm{\mu }}_0\mathrm{I}\text{'}}{\mathrm{\pi r}}={\mathrm{B}}_{\mathrm{P}\mathrm{at}\mathrm{N}}=2\frac{{\mathrm{\mu }}_0\mathrm{I}}{\mathrm{\pi r}} \\ \mathrm{I}\text{'}=3\mathrm{I} \\ \mathrm{Now},\mathrm{the}\mathrm{direction}\mathrm{of}\mathrm{the}\mathrm{current}\mathrm{in}\mathrm{Q}\mathrm{should}\mathrm{be}\mathrm{opposite}\mathrm{ot}\mathrm{the}\mathrm{direction}\mathrm{of}\mathrm{the}\mathrm{current}\mathrm{in}\mathrm{P} \\ \mathrm{as}\mathrm{per}\mathrm{the}\mathrm{right}\mathrm{hand}\mathrm{thumb}\mathrm{rule}\mathrm{to}\mathrm{create}\mathrm{the}\mathrm{desired}\mathrm{effect}. \end{gathered}$$