Book edition 10th Edition

Author(s) David Halliday

Pages 1328 pages

ISBN 9781118230718

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Short Answer

In Fig. 29-4, a wire forms a semicircle of radius $R = 9.26 cm$ and two (radial) straight segments each of length $L = 13.1 cm$ . The wire carries current $i = 34.8 mA$ . What are the(a) magnitude and(b) direction (into or out of the page) of the net magnetic field at the semicircle’s center of curvature C?

The magnitude of the magnetic field is $B = 1.18 \times 10^{- 7} T$
The direction of the magnetic field is into the page.

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Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: Given

Radius of circle is $R = 9.26 cm = 9.26 \times 10^{- 2} m$
Length of each straight segment is role="math" localid="1663098315135" $L = 13.1 cm = 13.1 \times 10^{- 2} m$
Current is $i = 34.8 mA = 34.8 \times 10^{- 3} A$

Step 2: Determine the formulas

The magnitude of the magnetic field B at centre of circular arc, of radius R, and central angle $ϕ$ , carrying a current I is given as follows:

$B = \frac{μ I ϕ}{4 π R}$

For a straight conductor consider the formulas:

$d \vec{B} = \frac{μ_{0} I d \vec{s} \times \hat{r}}{4 π r^{2}}$

Here, $d \bar{B}$ is the magnetic field through the current carrying conductor, $d \bar{s}$ length of the element and $\hat{r}$ is unit vector that is a point from the element to the given point.

Formulae:

$B = \frac{μIϕ}{4 πR} B = \frac{μIds sinθ}{4 {πr}^{2}}$

Step 3: (a) Calculate the magnitude of the net magnetic field at the semicircle’s center of curvature C

To calculate the magnetic field $B_{1}$ due to current for left straight segment as follows:

$B_{1} = \frac{μ I d s s i n θ}{4 π r^{2}}$

$θ = 0 °$

So,

$B_{1} = 0$

Calculate the magnetic field $B_{2}$ due to current for a right straight segment as follows:

$B_{2} = \frac{μ I d s s i n θ}{4 π r^{2}}$

Here $θ = 0 °$

$B_{2} = 0$

Calculate the magnetic field $B_{3}$ due to current from circular section as follows

$B_{3} = \frac{μ I ϕ}{4 π R} B_{3} = \frac{(4 π \times 10^{- 7} {N/A}^{2}) \times 34.8 \times 10^{- 3} A \times π}{4 π \times 9.26 \times 10^{- 2} m} B_{3} = 1.18 \times 10^{- 7} T$

So, the total magnetic field at the center of semicircle arc is

$B = B_{1} + B_{2} + B_{3} = 1.18 \times 10^{- 7} T$

Hence the magnetic field is, $1.18 \times 10^{- 7} T$ .

Step 4: (b) Calculate the direction (into or out of the page) of the net magnetic field at the semicircle’s center of curvature C

Using the right hand rule, direction of magnetic field is into the page

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Fundamentals Of Physics

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Short Answer

Step by Step Solution

Step 1: Given

Step 2: Determine the formulas

Step 3: (a) Calculate the magnitude of the net magnetic field at the semicircle’s center of curvature C

Step 4: (b) Calculate the direction (into or out of the page) of the net magnetic field at the semicircle’s center of curvature C

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Fundamentals Of Physics

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Short Answer

In Fig. 29-4, a wire forms a semicircle of radius R=9.26 cm and two (radial) straight segments each of length L=13.1 cm. The wire carries current i=34.8 mA. What are the(a) magnitude and(b) direction (into or out of the page) of the net magnetic field at the semicircle’s center of curvature C?

Step by Step Solution

Step 1: Given

Step 2: Determine the formulas

Step 3: (a) Calculate the magnitude of the net magnetic field at the semicircle’s center of curvature C

Step 4: (b) Calculate the direction (into or out of the page) of the net magnetic field at the semicircle’s center of curvature C

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