Book edition 10th Edition

Author(s) David Halliday

Pages 1328 pages

ISBN 9781118230718

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Short Answer

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 $i_{1} = 2.0 A$ in the direction indicated. Wire 2 is long and straight; it carries a current i₂ that can be varied; and it is at distance $\frac{R}{2}$ from the center of the arc. The net magnetic field $\vec{B}$ 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 $i_{2 s} = 1.00 A$ . What is the angle subtended by the arc?

The angle subtended by the arc is $θ = 1.00 rad$ .

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Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: Given

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

ii) The horizontal scale is set by $i_{2 x} = 1.00 A$

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:

$B_{s t r a i g h t} = \frac{μ_{0} i}{2 π R}$

$B_{a r c} = \frac{μ_{0} i \emptyset}{4 π R}$

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

$B_{s t r a i g h t} = \frac{μ_{0} i}{2 π R}$

Here, from the figure,

$B_{s t r a i g h t} = \frac{μ_{0} i_{2}}{2 π (\frac{R}{2})}$

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

$B_{a r c} = \frac{μ_{0} i_{1} \emptyset}{4 π R}$

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

$B = B_{a r c} - B_{s t r a i g h t}$

$B = \frac{μ_{0} i_{1} \emptyset}{4 π R} - \frac{μ_{0} i_{2}}{2 π (\frac{R}{2})}$ …… (1)

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

Hence, equation (1) becomes

$0 = \frac{μ_{0} i_{1} \emptyset}{4 π R} - \frac{μ_{0} i_{2}}{2 π (\frac{R}{2})}$

$\frac{μ_{0} i_{1} \emptyset}{4 π R} = \frac{μ_{0} i_{2}}{2 π (\frac{R}{2})}$

Solve further as:

role="math" localid="1663153369320" $\emptyset = 4 (\frac{i_{2}}{i_{1}})$

$\emptyset = 4 (\frac{0.50 A}{2.00 A})$

$\emptyset = 1.00 rad$

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