Book edition 1st Edition

Author(s) Paul Peter Urone

Pages 1272 pages

ISBN 9781938168000

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Short Answer

Find the magnitude and direction of the magnetic field at the point equidistant from the wires in Figure 22.58 (a)using the rules of vector addition to sum the contributions from each wire.

The magnitude of the magnetic field at the point equidistant from the wires is $7.55 \times 10^{- 5} T$ and its direction is $23 . 4^{°}$ from the horizontal axis.

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Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: Definition of magnetic field

A magnetic field is defined as a position in space near a magnet or an electric current where a physical field is formed by a moving electric charge applying force on another moving electric charge.

Step 2: Formula to used

The expression to calculate the magnetic field is,

$B = \frac{{{\mu _0}I}}{{2\pi r}}$

Where ${\mu _o} = $ is the magnetic permeability of the free space.

Step 3: The required magnitude and the direction of the magnetic field at the point equidistant from the wires.

The given figure can be drawn as given below.

In the diagram above, O represents the point that is equidistant from the wires.

$\begin{align}{}\frac{d}{{5\;cm}} &= \tan {30^\circ }\\d &= 2.88 \times {10^{ - 2}}\;m\end{align}$

Use Pythagoras theorem,

$\begin{align}{}r &= \sqrt {{{\left( {5.00 \times {{10}^{ - 2}}\;m} \right)}^2} + {{\left( {2.88 \times {{10}^{ - 2}}\;m} \right)}^2}} \\r &= 5.77 \times {10^{ - 2}}\;m\end{align}$

The magnetic field produced by wire A at position O is,

$\begin{align}{}{B_1} & = \frac{{{\mu _0}{I_1}}}{{2\pi r}}\\{B_1} & = \frac{{\left( {4\pi \times {{10}^{ - 7}}\;T \cdot m/A} \right)(5.00\;A)}}{{2\pi \left( {5.77 \times {{10}^{ - 2}}\;m} \right)}}\\{B_1} & = 1.73 \times {10^{ - 5}}\;T\end{align}$

The magnetic field produced by wire B at position O is,

$\begin{align}{}{B_2} & = \frac{{{\mu _0}{I_2}}}{{2\pi r}}\\{B_2} & = \frac{{\left( {4\pi \times {{10}^{ - 7}}\;T \cdot m/A} \right)(10.0\;A)}}{{2\pi \left( {5.77 \times {{10}^{ - 2}}\;m} \right)}}\\{B_2} & = 3.46 \times {10^{ - 5}}\;T\end{align}$

The magnetic field produced by wire C at position O is,

$\begin{align}{}{B_3} & = \frac{{{\mu _0}{I_3}}}{{2\pi r}}\\{B_3} & = \frac{{\left( {4\pi \times {{10}^{ - 7}}\;T \cdot m/A} \right)(20.0\;A)}}{{2\pi \left( {5.77 \times {{10}^{ - 2}}\;m} \right)}}\\{B_3} & = 6.93 \times {10^{ - 5}}\;T\end{align}$

Draw a vector of magnetic field

Along the horizontal axis, the overall magnetic is,

$\begin{align}{}{B_x} &= {B_1} + {B_2}\cos {60^\circ } + {B_3}\cos {60^\circ }{B_x}\\ &= 1.733 \times {10^{ - 5}}\;T + \left( {3.46 \times {{10}^{ - 5}}\;T} \right)(0.5) + \left( {6.932 \times {{10}^{ - 5}}\;T} \right)(0.5){B_x}\\ &= 6.93 \times {10^{ - 5}}\;T\end{align}$

Along the vertical axis, the overall magnetic is,

$\begin{align}{}{B_y} &= {B_3}\sin {60^\circ } + {B_2}\sin {60^\circ }{B_y}\\ &= \left( {6.932 \times {{10}^{ - 5}}\;T} \right)(0.866) - \left( {3.46 \times {{10}^{ - 5}}\;T} \right)(0.866){B_y}\\ &= 3.00 \times {10^{ - 5}}\;T\end{align}$

The resultant magnetic field is shown in the diagram below.

The resultant magnetic field is given as,

$\begin{align}{}B &= \sqrt {B_x^2 + B_y^2} \\ &= \sqrt {{{\left( {6.932 \times {{10}^{ - 5}}\;T} \right)}^2} + {{\left( {3.00 \times {{10}^{ - 5}}\;T} \right)}^2}} \\B &= 7.55 \times {10^{ - 5}}\;T\end{align}$

The direction of resultant magnetic field is,

$\begin{align}{}\tan \theta & = \frac{{3.00 \times {{10}^{ - 5}}\;T}}{{6.932 \times {{10}^{ - 5}}\;T}}\\\theta & = {23.4^\circ }\end{align}$

Therefore, the magnitude value is $7.55 \times {10^{ - 5}}\;T$and its angle value is ${23.4^\circ }$.

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College Physics (Urone)

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Short Answer

Find the magnitude and direction of the magnetic field at the point equidistant from the wires in Figure 22.58 (a)using the rules of vector addition to sum the contributions from each wire.

Step by Step Solution

Step 1: Definition of magnetic field

Step 2: Formula to used

Step 3: The required magnitude and the direction of the magnetic field at the point equidistant from the wires.

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College Physics (Urone)

Answers without the blur.

Short Answer

Find the magnitude and direction of the magnetic field at the point equidistant from the wires in Figure 22.58 (a)using the rules of vector addition to sum the contributions from each wire.

Step by Step Solution

Step 1: Definition of magnetic field

Step 2: Formula to used

Step 3: The required magnitude and the direction of the magnetic field at the point equidistant from the wires.

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