Figure 29-62 shows, in cross section, two long straight wires held against a plastic cylinder of radius$\mathbf{20}\mathbf{.}\mathbf{0}\mathit{c}\mathit{m}$. Wire 1 carries currentrole="math" localid="1663151041166" ${\mathit{i}}_{\mathbf{1}}\mathbf{=}\mathbf{60}\mathbf{.}\mathbf{0}\mathit{m}\mathit{A}$out of the page and is fixed in place at the left side of the cylinder. Wire 2 carries current${\mathit{i}}_{\mathbf{2}}\mathbf{=}\mathbf{40}\mathbf{.}\mathbf{0}\mathit{m}\mathit{A}$out of the page and can be moved around the cylinder. At what (positive) angle${\mathbf{\theta }}_{\mathbf{2}}$should wire 2 be positioned such that, at the origin, the net magnetic field due to the two currents has magnitude$\mathbf{80}\mathbf{.}\mathbf{0}\mathbf{nT}$?

1. The radius of the plastic cylinder $R=20.0cmor0.200m$
2. Wire 1 carries current $i_1=60.0mAor60.0\times {10}^{-3}A$
3. Wire 2 carries current $i_2=40.0mAor40.0\times {10}^{-3}A$
4. The net magnetic field due to$i_{1}and{i}_{2}isB=80.0nTor80\times {10}^{-9}T$

When two parallel current-carrying conductors are placed near each other, they exert a force on each other, either attractive or repulsive, depending on the direction of the current. The fields are expressed by Biot-Savart’s law. It is stated as,

$\mathit{d}\overrightarrow{\mathbf{B}}\mathbf{=}\frac{{\mathbf{\mu }}_{\mathbf{0}}}{\mathbf{4}\mathbf{\pi }}\frac{\mathbf{i}\mathbf{d}\overrightarrow{\mathbf{s}}\mathbf{\times }\hat{\mathbf{r}}}{{\mathbf{r}}^{\mathbf{2}}}$

Here, $\mathit{d}\mathit{B}$is the field produced by the current element $\mathit{i}\mathit{d}\overrightarrow{\mathbf{s}}$at a point located at $\mathit{r}$from the current element.

Using Biot-Savart’s law and Lorentz force equation, we can calculate the force on the current carrying conductor by another current carrying conductor.

By using, $\mathbf{\text{Equation}}\mathbf{}\mathbf{29}\mathbf{-}\mathbf{4}$we can find the magnitude of themagnetic fields.Also, we have to find the net components${\mathit{B}}_{\mathbf{2}\mathbf{x}}$and${\mathit{B}}_{\mathbf{2}\mathbf{y}}$of${\mathit{B}}_{\mathbf{2}}$. Now, by usingPythagoras theorem and substituting the given quantities, we can find the angle${\mathit{\theta }}_{\mathbf{2}}$.

Formulas:

From , $29-4$the magnitude of the magnetic field is given by

$B=\frac{{\mu }_{0}i}{2\mathrm{\pi R}}$

Here, $B$is magnetic field, ${\mu }_0$is permeability constant, $i$is current, $R$is the distance between the point and the current conducting wire.

From,$\text{equation}29-4$the magnitude of the magnetic field is given by

$B=\frac{{\mu }_{0}i}{2\pi R}$

By the right-hand rule, the field caused by wirecurrent, evaluated at the coordinate origin, is along$+y$axis.

The field caused by the current of wire$2\text{'}s$ will generally have both an $x$and a$y$ component which are related to its magnitude$B_2$and sines and cosines of some angle. When wire 2 is at an angle$\theta$, its net components are given by

$B_{2x}=B_{2}sin{\theta }_{2}and{B}_{2y}=-B_{2}cos{\theta }_2$

By Pythagoras theorem, the magnitude-squared of the net field is equal to the sum of the square of their net x-component and square of their net y-component.

Therefore,

$$\begin{gathered} B^2={\left(B_2\sin {\theta }_2\right)}^2+{\left(B_1-B_2\cos {\theta }_2\right)}^2 \\ =B_1^2+B_2^2{\sin }^2\theta +B_2^2{\cos }^2\theta -2B_1{B}_2\cos {\theta }_2 \end{gathered}$$

$B^2=B_1^2+B_2^2{\sin }^2\theta +B_2^2{\cos }^2\theta -2B_1{B}_{2}cos{\theta }_2$

Since,${\sin }^2\theta +{\cos }^2\theta =1$

$B^2=B_1^2+B_2^2-2B_1{B}_{2}cos{\theta }_2$

Therefore, angle${\theta }_2$is

${\theta }_2={\cos }^{-1}\left[\frac{\left(B_1^2+B_2^2-B^2\right)}{2B_1{B}_2}\right]$

Now, we have

$$\begin{gathered} {\mathrm{B}}_1=\frac{{\mathrm{\mu }}_0{\mathrm{i}}_1}{2\mathrm{\pi R}} \\ =\frac{\left(4\mathrm{\pi }\times {10}^{-7}\mathrm{T}·\mathrm{m}/\mathrm{A}\right)\times \left(60.0\times {10}^{-3}\mathrm{A}\right)}{2\times \mathrm{\pi }\times 0.20\mathrm{m}} \\ =60\times {10}^{-9}\mathrm{T} \end{gathered}$$

Similarly,

$$\begin{gathered} B_2=\frac{{\mu }_{0i_2}}{2\mathrm{\pi R}} \\ =\frac{\left(4\mathrm{\pi }\times {10}^{-7}\mathrm{T}·\mathrm{m}/\mathrm{A}\right)\times \left(40.0\times {10}^{-3}A\right)}{2\times \mathrm{\pi }\times 0.20\mathrm{m}} \\ =40\times {10}^{-9}T \end{gathered}$$

With the requirement that the net field has magnitude$B=80\text{nT}$, we find

$$\begin{gathered} {\theta }_2={\cos }^{-1}\left[\frac{\left(B_1^2+B_2^2-B^2\right)}{2B_1{B}_2}\right] \\ ={\text{cos}}^{-1}\left[\frac{{\left(60\times {10}^{-9} \text{T}\right)}^2+{\left(40\times {10}^{-9} \text{T}\right)}^2-{\left(80\times {10}^{-9} \text{T}\right)}^2}{2\times \left(60\times {10}^{-9} \text{T}\right)\left(40\times {10}^{-9} \text{T}\right)}\right] \\ ={\text{cos}}^{-1}\left[-0.25\right] \\ =104° \end{gathered}$$

Therefore, the angle ${\theta }_{2}is104°.$