91Ó°ÊÓ

• Direction of current through wire 1 is coming out of plane of figure
• Direction of current through wire 2 is going into plane of figure
• $i=4.23\text{A}$ ,for both wire$d=18.6\mathrm{cm}=18.6×{10}^{-2}\mathrm{m}$
• Distance of point P from origin is$=34.2\text{cm}=34.2×{10}^{-2}\text{m}$

For this problem, we need to superpose the two fields at point p due to long straight wires 1 and 2. By using Pythagoras theorem, we can find the distance between 1 and p , also between 2 and p. Then we can use the formula for magnetic field due to long straight wire at a pointP at a distance d from the wire carrying current i for both wires. We can add them as vectors to find the net field.

Formula:

$B_z=\frac{{\mathrm{μ}}_{0}i}{2\mathrm{π}z}$

Let${\stackrel{→}{B}}_1$be the magnetic field at point P due wirewhich lies in first quadrant ofplane xy and makes angle$\left(\mathrm{θ}-\frac{\mathrm{π}}{2}\right)$with x axis; in this case, we assume that the angle made by the line from wire 1 to pointwith x axis. Hence we get

${\stackrel{→}{B}}_1=\frac{{\mathrm{μ}}_{0}i}{2\mathrm{Ï€}z}·\text{cos}\left(\mathrm{θ}-\frac{\mathrm{Ï€}}{2}\right).\stackrel{â‡\P Ä}{\mathrm{i}}+\frac{{\mathrm{μ}}_{0}i}{2\mathrm{Ï€}z}.\sin \left(\mathrm{θ}-\frac{\mathrm{Ï€}}{2}\right).\stackrel{â‡\P Ä}{\mathrm{j}}$

Let be the magnetic field at point P due wire which lies in the first quadrant of xy plane and makes angle $\left(\mathrm{θ}-\frac{\mathrm{π}}{2}\right)$ with x axis. In this case, we assume that the angle made by the line from wire 2 to the point P is $\mathrm{θ}$with x axis. Hence we get

${\stackrel{→}{B}}_2=\frac{{\mathrm{μ}}_{0}i}{2\mathrm{Ï€}z}·\text{cos}\left(\mathrm{θ}-\frac{\mathrm{Ï€}}{2}\right).\stackrel{â‡\P Ä}{\mathrm{i}}-\frac{{\mathrm{μ}}_{0}i}{2\mathrm{Ï€}z}.\sin \left(\mathrm{θ}-\frac{\mathrm{Ï€}}{2}\right).\stackrel{â‡\P Ä}{\mathrm{j}}$

Hence, the net field is given as follows

. $\begin{array}{rcl}{\stackrel{→}{B}}_{net}& =& \frac{{\mathrm{μ}}_{0}id}{2\mathrm{Ï€}{z}^2}\stackrel{â‡\P Ä}{\mathrm{i}}=\frac{{\mathrm{μ}}_{0}id}{2\mathrm{Ï€}\left(R^2+\frac{d^2}{4}\right)}\stackrel{â‡\P Ä}{\mathrm{i}}=\frac{2{\mathrm{μ}}_{0}id}{4\mathrm{Ï€}\left(R^2+\frac{d^2}{4}\right)}\stackrel{â‡\P Ä}{\mathrm{i}}\\ {\stackrel{→}{B}}_{net}& =& \frac{(2×{10}^{-7}×4.23×18.6×{10}^{-2})}{{\left(34.2×{10}^{-2}\right)}^2+\frac{{\left(18.6×{10}^{-2}\right)}^2}{4}}\stackrel{â‡\P Ä}{\mathrm{i}}\\ {\stackrel{→}{B}}_{net}& =& \left(1.25×{10}^{-6}\text{T}\right)\stackrel{â‡\P Ä}{\mathrm{i}}\end{array}$

Hence, the net magnetic field at point P due to two currents in wire 1 and wire 2, ${\stackrel{→}{B}}_{net}$is ${\stackrel{→}{B}}_{net}=\left(1.25×{10}^{-6}\text{T}\right)$