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P 27

Expert-verifiedFound in: Page 836

Book edition
10th Edition

Author(s)
David Halliday

Pages
1328 pages

ISBN
9781118230718

**Question: In Fig 29-55, two long straight wires (shown in cross section) carry currents** ${{\mathbf{i}}}_{{\mathbf{1}}}{\mathbf{=}}{\mathbf{30}}{\mathbf{.}}{\mathbf{0}}{\mathbf{}}{\mathbf{mA}}$** and **${{\mathbf{i}}}_{{\mathbf{1}}}{\mathbf{=}}{\mathbf{40}}{\mathbf{.}}{\mathbf{0}}{\mathbf{}}{\mathbf{mA}}$** directly out of the page. They are equal distances from the origin, where they set up a magnetic field. To what value must current** ${{\mathbf{i}}}_{{\mathbf{1}}}$** be changed in order to rotate** ${\mathbf{20}}{\mathbf{.}}{\mathbf{0}}{\mathbf{\xb0}}$** clockwise?**

The value of current ${i}_{1}$ is ${i}_{1}=61.3mA$.

i) Currents flowing through the two long straight wires are ${i}_{1}=30.0mA$ and ${i}_{2}=40.0mA$

ii) The rotation of net magnetic field $\overrightarrow{B}$ is $\theta =20.0\xb0$.

Use the concept of the magnetic force due to current in straight wires and trigonometry.

**Formulae:**

**${B}_{straight}=\frac{{\mu}_{0}i}{4\pi R}$**

$\mathrm{tan}\theta =\frac{{B}_{y}}{{B}_{x}}$

The value of current ${i}_{1}$:

The magnetic field due to a current in straight wire is

${B}_{straight}=\frac{{\mu}_{0}i}{4\pi R}$

The distances of the ${B}_{1}$ and ${B}_{2}$ are the same; hence they are directly proportional localid="1663143974221" ${i}_{1}$ and ${i}_{2}$ respectively.

${B}_{1}\alpha {i}_{1}$ and

${B}_{2}\alpha {i}_{2}$

According to the right hand rule, is going to the y axis and is going along x axis.

The angle of the net field is

$\mathrm{tan}\theta =\frac{{B}_{y}}{{B}_{x}}$

$\mathrm{tan}\theta =\frac{{B}_{2}}{{B}_{1}}$

$\theta ={\mathrm{tan}}^{-1}\left(\frac{{i}_{2}}{{i}_{1}}\right)$

Substitute the values and solve as:

$\theta ={\mathrm{tan}}^{-1}\left(\frac{40.0\text{mA}}{30.0\text{mA}}\right)$

$\theta =53.13\xb0$

In the problem, the net field rotation is

$\theta \text{'}=\theta -20.0\xb0$

$\theta \text{'}=53.13\xb0-20.0\xb0$

$\theta \text{'}=33.13\xb0$

The final value of the current is:

$\mathrm{tan}\theta \text{'}=\frac{{i}_{2}}{{i}_{1}}$

${i}_{1}=\frac{{i}_{2}}{\mathrm{tan}\theta \text{'}}$

Substitute the values and solve as:

${i}_{1}=\frac{40.0\text{mA}}{\mathrm{tan}\left(33.13\xb0\right)}$

${i}_{1}=61.3\text{mA}$

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