Figure 29-85 shows, in cross section, two long parallel wires that are separated by distance $\mathit{d}\mathbf{=}\mathbf{18}\mathbf{.}\mathbf{6}\mathbf{}\mathbf{\text{cm}}$. Each carries $\mathbf{4}\mathbf{.}\mathbf{23}\mathbf{}\mathbf{\text{A}}$, out of the page in wire 1 and into the page in wire 2. In unit-vector notation, what is the net magnetic field at point Pat distance $\mathit{R}\mathbf{=}\mathbf{34}\mathbf{.}\mathbf{2}\mathbf{}\mathbf{\text{cm}}$, due to the two currents?

• 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$d=18.6\mathrm{cm}=18.6\times {10}^{-2}\mathrm{m}$
• Distance of point P from origin is$=34.2\text{cm}=34.2\times {10}^{-2}\text{m}$

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:

$B_z=\frac{{\mu }_{0}i}{2\pi z}$

Let${\overrightarrow{B}}_1$be the magnetic field at point P due wirewhich lies in first quadrant ofplane xy and makes angle$\left(\theta -\frac{\pi }{2}\right)$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

${\overrightarrow{B}}_1=\frac{{\mu }_{0}i}{2\pi z}·\text{cos}\left(\theta -\frac{\pi }{2}\right).\stackrel{\rightharpoonup }{\mathrm{i}}+\frac{{\mu }_{0}i}{2\pi z}.\sin \left(\theta -\frac{\pi }{2}\right).\stackrel{\rightharpoonup }{\mathrm{j}}$

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

${\overrightarrow{B}}_2=\frac{{\mu }_{0}i}{2\pi z}·\text{cos}\left(\theta -\frac{\pi }{2}\right).\stackrel{\rightharpoonup }{\mathrm{i}}-\frac{{\mu }_{0}i}{2\pi z}.\sin \left(\theta -\frac{\pi }{2}\right).\stackrel{\rightharpoonup }{\mathrm{j}}$

Hence, the net field is given as follows

. $\begin{array}{rcl}{\overrightarrow{B}}_{net}& =& \frac{{\mu }_{0}id}{2\pi z^2}\stackrel{\rightharpoonup }{\mathrm{i}}=\frac{{\mu }_{0}id}{2\pi \left(R^2+\frac{d^2}{4}\right)}\stackrel{\rightharpoonup }{\mathrm{i}}=\frac{2{\mu }_{0}id}{4\pi \left(R^2+\frac{d^2}{4}\right)}\stackrel{\rightharpoonup }{\mathrm{i}}\\ {\overrightarrow{B}}_{net}& =& \frac{(2\times {10}^{-7}\times 4.23\times 18.6\times {10}^{-2})}{{\left(34.2\times {10}^{-2}\right)}^2+\frac{{\left(18.6\times {10}^{-2}\right)}^2}{4}}\stackrel{\rightharpoonup }{\mathrm{i}}\\ {\overrightarrow{B}}_{net}& =& \left(1.25\times {10}^{-6}\text{T}\right)\stackrel{\rightharpoonup }{\mathrm{i}}\end{array}$

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