Chapter 19, Problem 69P

To determine

The magnetic field at point P.

Answer to Problem 69P

The magnetic field at point P is $\underset{\_}{1.6\times {10}^{-5}\text{T out of the page}}$.

Explanation of Solution

Write the equation of net magnetic field.

  $B=B_{\text{bottom}}+B_{\text{top}}$                                                                                                 (I)

Here, $B$ is the net magnitude field, $B_{\text{bottom}}$ is the magnetic field due to the bottom wire and $B_{\text{top}}$ is the magnetic field due to the top wire.

Write the equation of magnetic field from bottom wire.

  $B_{\text{bottom}}=\frac{{\mu }_{0}I}{2\pi r_{\text{bottom}}}$                                                                                             (II)

Here, ${\mu }_0$ is the permeability of free space, $I$ is the current and $r_{\text{bottom}}$ is the distance from bottom wire.

Write the equation of magnetic field from top wire.

  $B_{\text{top}}=\frac{{\mu }_{0}I}{2\pi r_{\text{top}}}$                                                                                                 (III)

Here, $r_{\text{top}}$ is the distance from top wire.

Write the equation of distance from top wire.

  $r_{\text{top}}=r_{\text{bottom}}+r_1$                                                                                                 (IV)

Here, $r_1$ is the distance.

Rewrite the expression (I).

  $\begin{array}{c}B=\left(\frac{{\mu }_{0}I}{2\pi r_{\text{bottom}}}\right)+\left(\frac{{\mu }_{0}I}{2\pi r_{\text{top}}}\right)\\ =\frac{{\mu }_{0}I}{2\pi }\left[\left(\frac{1}{r_{\text{bottom}}}\right)+\left(\frac{1}{r_{\text{bottom}}+r_1}\right)\right]\end{array}$                                                                   (V)

Conclusion:

Substitute $4\pi \times {10}^{-7}\text{T-m/A}$ for ${\mu }_0$, $10.0\text{A}$ for $I$, $25\text{cm}$ for $r_{\text{bottom}}$ and $3.0\text{mm}$ for $r_1$ in equation (V) to find $B$.

  $\begin{array}{c}B=\frac{\left(4\pi \times {10}^{-7}\text{T-m/A}\right)\left(10.0\text{A}\right)}{2\pi }\left(\begin{array}{l}\left(\frac{1}{\left[\left(25\text{cm}\right)\left(\frac{1\times {10}^{-2}\text{m}}{1\text{cm}}\right)\right]}\right)+\\ \left(\frac{1}{\left(\left(25\text{cm}\right)\left(\frac{1\times {10}^{-2}\text{m}}{1\text{cm}}\right)\right)+\left(\left(3.0\text{mm}\right)\left(\frac{1\times {10}^{-3}\text{m}}{1\text{mm}}\right)\right)}\right)\end{array}\right)\\ =\frac{\left(4\pi \times {10}^{-7}\text{T-m/A}\right)\left(10.0\text{A}\right)}{2\pi }\left[\begin{array}{l}\left(\frac{1}{\left(0.25\text{m}\right)}\right)+\\ \left(\frac{1}{\left(0.25\text{m}\right)+\left(0.003\text{m}\right)}\right)\end{array}\right]\\ =1.6\times {10}^{-5}\text{T}\end{array}$

The direction of the magnetic field directed along out of the page.

Thus, the magnetic field at point P is $\underset{\_}{1.6\times {10}^{-5}\text{T out of the page}}$.