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P 30

Expert-verifiedFound in: Page 836

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 ${\mathit{R}}{\mathbf{=}}{\mathbf{20}}{\mathbf{.}}{\mathbf{0}}{\mathbf{}}{\mathbf{}}{\mathbf{\text{cm}}}$**** from the cylinder’s central axis.**

**Figure 29-58 a 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" ${{\mathit{\theta}}}_{{\mathbf{1}}}{\mathbf{=}}{\mathbf{}}{\mathbf{0}}{\mathbf{\xb0}}$ to angle localid="1663154390159" ${{\mathit{\theta}}}_{{\mathbf{1}}}{\mathbf{=}}{\mathbf{180}}{\mathbf{\xb0}}$**

a) The angle ${\theta}_{2}$ at which wire 2 is located is either $+90\xb0\text{or}-90\xb0$.

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.

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:

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

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

$\overrightarrow{B}=\overrightarrow{{B}_{1}}+\overrightarrow{{B}_{2}}$

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 $90\xb0$ or (i.e., $270\xb0$)

To avoid the collision of two wires, wire 2 must be at either $+90\xb0\text{or}-90\xb0$.

Hence, the angle ${\theta}_{2}$ at which wire 2 is located is either $+90\xb0\text{or}-90\xb0$.

We have

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

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

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 $180\xb0$. 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

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

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

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 $-90\xb0$, the current must be going into the page

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

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