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Q69PE

Expert-verifiedFound in: Page 814

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 $7.55\times {10}^{-5}\mathrm{T}$ and its direction is $23.{4}^{\xb0}$ from the horizontal axis.

**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.**

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.

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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