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Expert-verified Found in: Page 814 ### College Physics (Urone)

Book edition 1st Edition
Author(s) Paul Peter Urone
Pages 1272 pages
ISBN 9781938168000 # 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 $7.55×{10}^{-5}\mathrm{T}$ and its direction is $23.{4}^{°}$ from the horizontal axis.

See the step by step solution

## 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 $${23.4^\circ }$$. ### Want to see more solutions like these? 