91Ó°ÊÓ

Thedistance between the center of the cylinder and the wires is$R=20.0cm$.

The wire 2 is fixed in place.

The magnetic field at a point due to a current carrying wire is determined using the Biot-Savart law. The direction of the magnetic field is decided by the right-hand rule. The net magnetic field at the center of the cylinder is the vector sum of the magnetic fields due to wire 1 and wire 2. Using this, we can answer the above questions.

Formula:

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

The information about the net magnetic field at the center of the cylinder is given in figures (b) and (c). Using this, we can write the equations as

$\stackrel{→}{B}=\stackrel{→}{B_1}+\stackrel{→}{B_2}$

In the component form,

$$\begin{gathered} B_x=B_{1x}+B_{2x} \\ {B}_y=B_{1y}+B_{2y} \end{gathered}$$

$$\begin{gathered} B_{x,0}=B_{1x,0}+B_{2x} \\ {B}_{y,0}=B_{1y,0}+B_{2y} \\ {B}_{x,90}=B_{1x,90}+B_{2x} \\ {B}_{y,90}=B_{1y,90}+B_{2y} \\ {B}_{x,180}=B_{1x,180}+B_{2x} \\ {B}_{y,180}=B_{1y,180}+B_{2y} \end{gathered}$$

Using the values form the graph, we write

data-custom-editor="chemistry" $$\begin{gathered} 2=0+B_{2x} \\ -4=B_{1y,0}+B_{2y} \\ 6=B_{1x,90}+B_{2x} \\ 0=0+B_{2y} \\ 2=0+B_{2x} \\ 4=B_{1y,180}+B_{2y} \end{gathered}$$

From these equations, we can find

$B_{2x}=2\mathrm{μ}\text{T}$and$B_{2y}=0$

Hence the wire 2 must be along the y axis direction, i.e., its position must be at$90°$or (i.e.,$270°$)

To avoid the collision of two wires, wire 2 must be at either $+90°\text{or}-90°$.

Hence, the angle ${\mathrm{θ}}_2$at which wire 2 is located is either$+90°\text{or}-90°$.

We have

$$\begin{gathered} B_{1y}=\frac{{\mathrm{μ}}_0{i}_1}{2\mathrm{π}R} \\ 4×{10}^{-6}=\frac{{\mathrm{μ}}_0{i}_1}{2\mathrm{π}R} \\ {B}_{1y,180}=4\mathrm{μ}\text{T} \end{gathered}$$

$\begin{array}{rcl}{i}_1& =& \frac{4×{10}^{-6}×2\mathrm{π}×20×{10}^{-2}}{4\mathrm{π}×{10}^{-7}}\\ i_1& =& 40×{10}^{-1}\\ & =& 4.0\text{A}\end{array}$

Hence, the size of the current in wire 1 = 4.0 A.

The y component of the magnetic field is positive when wire 1 is at$180°$. This is possible only if the current in the wire 1 is out of the page. All the other equations are also valid for this direction of the current.

Hence, the direction of the current in wire 1= out of the page

We have

$$\begin{gathered} B_{2x}=2\mathrm{μ}T \\ {B}_{2x}=\frac{{\mathrm{μ}}_0{i}_2}{2\mathrm{π}R} \\ 2×{10}^{-6}=\frac{{\mathrm{μ}}_0{i}_2}{2\mathrm{π}R} \end{gathered}$$

$\begin{array}{rcl}{i}_2& =& \frac{2×{10}^{-6}×2\mathrm{π}×20×{10}^{-2}}{4\mathrm{π}×{10}^{-7}}\\ i_2& =& 20×{10}^{-1}\\ & =& 2.0\text{A}\end{array}$

Hence, the size of the current in wire 2 = 2.0 A.

The x component of the current in wire 2 is positive. Hence as seen in part (a), if the wire is at $-90°$, the current must be going into the page

Hence, the direction of the current in wire 2 is into the page.