91Ó°ÊÓ

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

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,

$\mathit{d}\stackrel{\mathbf{→}}{\mathbf{B}}\mathbf{=}\frac{{\mathbf{μ}}_{\mathbf{0}}}{\mathbf{4}\mathbf{π}}\frac{\mathbf{i}\mathbf{d}\stackrel{\mathbf{→}}{\mathbf{s}}\mathbf{×}\hat{\mathbf{r}}}{{\mathbf{r}}^{\mathbf{2}}}$

Here, $\mathit{d}\mathit{B}$is the field produced by the current element $\mathit{i}\mathit{d}\stackrel{\mathbf{→}}{\mathbf{s}}$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${\mathit{θ}}_{\mathbf{2}}$.

Formulas:

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

$B=\frac{{\mathrm{μ}}_{0}i}{2\mathrm{Ï€¸é}}$

Here, $B$is magnetic field, ${\mathrm{μ}}_0$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

$B=\frac{{\mathrm{μ}}_{0}i}{2\mathrm{π}R}$

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$\mathrm{θ}$, its net components are given by

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

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,

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

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

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

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

Therefore, angle${\mathrm{θ}}_2$is

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

Now, we have

$$\begin{gathered} {\mathrm{B}}_1=\frac{{\mathrm{μ}}_0{\mathrm{i}}_1}{2\mathrm{Ï€¸é}} \\ =\frac{\left(4\mathrm{Ï€}×{10}^{-7}\mathrm{T}·\mathrm{m}/\mathrm{A}\right)×\left(60.0×{10}^{-3}\mathrm{A}\right)}{2×\mathrm{Ï€}×0.20\mathrm{m}} \\ =60×{10}^{-9}\mathrm{T} \end{gathered}$$

Similarly,

$$\begin{gathered} B_2=\frac{{\mathrm{μ}}_{0i_2}}{2\mathrm{Ï€¸é}} \\ =\frac{\left(4\mathrm{Ï€}×{10}^{-7}\mathrm{T}·\mathrm{m}/\mathrm{A}\right)×\left(40.0×{10}^{-3}A\right)}{2×\mathrm{Ï€}×0.20\mathrm{m}} \\ =40×{10}^{-9}T \end{gathered}$$

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

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

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