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Found in: Page 836

### Fundamentals Of Physics

Book edition 10th Edition
Author(s) David Halliday
Pages 1328 pages
ISBN 9781118230718

# Question: Two long straight thin wires with current lie against an equally long plastic cylinder, at radius ${\mathbit{R}}{\mathbf{=}}{\mathbf{20}}{\mathbf{.}}{\mathbf{0}}{\mathbf{}}{\mathbf{}}{\mathbf{\text{cm}}}$ from the cylinder’s central axis.Figure 29-58a shows, in cross section, the cylinder and wire 1 but not wire 2. With wire 2 fixed in place, wire 1 is moved around the cylinder, from angle localid="1663154367897" ${{\mathbit{\theta }}}_{{\mathbf{1}}}{\mathbf{=}}{\mathbf{}}{\mathbf{0}}{\mathbf{°}}$ to angle localid="1663154390159" ${{\mathbit{\theta }}}_{{\mathbf{1}}}{\mathbf{=}}{\mathbf{180}}{\mathbf{°}}$, through the first and second quadrants of the xy coordinate system. The net magnetic field $\stackrel{\mathbf{\to }}{\mathbf{B}}$ at the center of the cylinder is measured as a function of ${\mathbf{}}{{\mathbit{\theta }}}_{{\mathbf{1}}}$. Figure 29-58b gives the x component ${\mathbf{}}{{\mathbit{B}}}_{{\mathbf{x}}}$ of that field as a function of ${\mathbf{}}{{\mathbit{\theta }}}_{{\mathbf{1}}}$(the vertical scale is set by ${{\mathbit{B}}}_{\mathbf{x}\mathbf{s}}{\mathbf{=}}{\mathbf{6}}{\mathbf{.}}{\mathbf{0}}{\mathbf{}}{\mathbf{}}{\mathbit{\mu }}{\mathbf{\text{T}}}$), and Fig. 29-58c gives the y component (the vertical scale is set by ${{\mathbit{B}}}_{\mathbf{y}\mathbf{s}}{\mathbf{}}{\mathbf{=}}{\mathbf{}}{\mathbf{4}}{\mathbf{.}}{\mathbf{0}}{\mathbf{}}{\mathbf{}}{\mathbit{\mu }}{\mathbf{\text{T}}}$). (a) At what angle ${\mathbf{}}{{\mathbit{\theta }}}_{{\mathbf{2}}}$ is wire 2 located? What are the (b) size and (c) direction (into or out of the page) of the current in wire 1 and the (d) size and (e) direction of the current in wire 2?

a) The angle ${\theta }_{2}$ at which wire 2 is located is either $+90°\text{or}-90°$.

b) The size of the current in wire 1 = 4.0 A.

c) The direction of the current in wire 1= out of the page

d) The size of the current in wire 2 = 2.0 A

e) The direction of the current in wire 2 is into the page.

See the step by step solution

## Step 1: The given values and the concept.

The distance 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:

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

## Step 2: (a) Calculate the angle  θ2 at which wire 2 is located.

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

$\stackrel{\to }{B}=\stackrel{\to }{{B}_{1}}+\stackrel{\to }{{B}_{2}}$

In the component form,

${B}_{x}={B}_{1x}+{B}_{2x}\phantom{\rule{0ex}{0ex}}{B}_{y}={B}_{1y}+{B}_{2y}$

${B}_{x,0}={B}_{1x,0}+{B}_{2x}\phantom{\rule{0ex}{0ex}}{B}_{y,0}={B}_{1y,0}+{B}_{2y}\phantom{\rule{0ex}{0ex}}{B}_{x,90}={B}_{1x,90}+{B}_{2x}\phantom{\rule{0ex}{0ex}}{B}_{y,90}={B}_{1y,90}+{B}_{2y}\phantom{\rule{0ex}{0ex}}{B}_{x,180}={B}_{1x,180}+{B}_{2x}\phantom{\rule{0ex}{0ex}}{B}_{y,180}={B}_{1y,180}+{B}_{2y}$

Using the values form the graph, we write

data-custom-editor="chemistry" $2=0+{B}_{2x}\phantom{\rule{0ex}{0ex}}-4={B}_{1y,0}+{B}_{2y}\phantom{\rule{0ex}{0ex}}6={B}_{1x,90}+{B}_{2x}\phantom{\rule{0ex}{0ex}}0=0+{B}_{2y}\phantom{\rule{0ex}{0ex}}2=0+{B}_{2x}\phantom{\rule{0ex}{0ex}}4={B}_{1y,180}+{B}_{2y}$

From these equations, we can find

${B}_{2x}=2\mu \text{T}$ 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 ${\theta }_{2}$ at which wire 2 is located is either $+90°\text{or}-90°$.

## Step 3: (b) Calculate the size of the current in wire 1

We have

${B}_{1y}=\frac{{\mu }_{0}{i}_{1}}{2\pi R}\phantom{\rule{0ex}{0ex}}4×{10}^{-6}=\frac{{\mu }_{0}{i}_{1}}{2\pi R}\phantom{\rule{0ex}{0ex}}{B}_{1y,180}=4\mu \text{T}$

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

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

## Step 4: (c) Calculate the direction of the current in wire 1

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

## Step 5: (d) Calculate the size of the current in wire 2

We have

${B}_{2x}=2\mu T\phantom{\rule{0ex}{0ex}}{B}_{2x}=\frac{{\mu }_{0}{i}_{2}}{2\pi R}\phantom{\rule{0ex}{0ex}}2×{10}^{-6}=\frac{{\mu }_{0}{i}_{2}}{2\pi R}$

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

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

## Step 6: (e) Calculate the direction of the current in wire 2

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.