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P 28

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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: Figure 29-56a shows two wires, each carrying a current .Wire 1 consists of a circular arc of radius R and two radial lengths; it carries current ${{\mathbit{i}}}_{{\mathbf{1}}}{\mathbf{=}}{\mathbf{}}{\mathbf{2}}{\mathbf{.}}{\mathbf{0}}{\mathbf{}}{\mathbf{\text{A}}}$ in the direction indicated. Wire 2 is long and straight; it carries a current i2 that can be varied; and it is at distance $\frac{\mathbf{R}}{\mathbf{2}}$ from the center of the arc. The net magnetic field $\stackrel{\mathbf{\to }}{\mathbf{B}\mathbf{}}$ due to the two currents is measured at the center of curvature of the arc. Figure 29-56b is a plot of the component of in the direction perpendicular to the figure as a function of current i2. The horizontal scale is set by ${{\mathbit{i}}}_{\mathbf{2}\mathbf{s}}{\mathbf{=}}{\mathbf{}}{\mathbf{1}}{\mathbf{.}}{\mathbf{00}}{\mathbf{}}{\mathbf{\text{A}}}$. What is the angle subtended by the arc?

The angle subtended by the arc is $\theta =1.00\text{rad}$.

See the step by step solution

Step 1: Given

i) The current flowing through the wire is ${i}_{1}=2.00A$.

ii) The horizontal scale is set by ${i}_{2x}=1.00\text{A}$

ii) The radius of the circular arc is R.

iv) The distance between wire 2 and the center of the circular arc of wire 1 is $\frac{R}{2}$.

Step 2: Understanding the concept

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

Formulae:

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

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

Step 3: Calculate the angle subtended by the arc

The angle subtended by the arc:

The magnetic field due to a current in long straight wire is

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

Here, from the figure,

${B}_{straight}=\frac{{\mu }_{0}{i}_{2}}{2\pi \left(\frac{R}{2}\right)}$

According to the right hand, both give direction of magnetic field pointing out of the page.

The magnetic field due to the current in circular arc of the wire is

${B}_{arc}=\frac{{\mu }_{0}{i}_{1}\varnothing }{4\pi R}$

In the figure (a), according to the right hand, both wires give the direction of magnetic field pointing out of the page.

The net magnetic field is

$B={B}_{arc}-{B}_{straight}$

$B=\frac{{\mu }_{0}{i}_{1}\varnothing }{4\pi R}-\frac{{\mu }_{0}{i}_{2}}{2\pi \left(\frac{R}{2}\right)}$ …… (1)

From the figure (b), for ${i}_{2x}=0.5\text{A,}$ then $B=0\text{T}$

Hence, equation (1) becomes

$0=\frac{{\mu }_{0}{i}_{1}\varnothing }{4\pi R}-\frac{{\mu }_{0}{i}_{2}}{2\pi \left(\frac{R}{2}\right)}$

$\frac{{\mu }_{0}{i}_{1}\varnothing }{4\pi R}=\frac{{\mu }_{0}{i}_{2}}{2\pi \left(\frac{R}{2}\right)}$

$\frac{{\mu }_{0}{i}_{1}\varnothing }{4\pi R}=\frac{{\mu }_{0}{i}_{2}}{2\pi \left(\frac{R}{2}\right)}$

Solve further as:

role="math" localid="1663153369320" $\varnothing =4\left(\frac{{i}_{2}}{{i}_{1}}\right)$

$\varnothing =4\left(\frac{0.50A}{2.00A}\right)$

$\varnothing =1.00\text{rad}$

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