Chapter 19, Problem 85P

To determine

The magnetic field at point R.

Answer to Problem 85P

The magnetic field at point R is $\underset{\_}{96\text{μT and directed to the right}}$.

Explanation of Solution

The diagram for the magnetic field is depicted in figure 1:

The vertical component of the two fields will cancel each other and the horizontal component will added up.

Write the equation of net magnetic field.

  $B_{net}=2B_{\text{left}}+2B_{\text{right}}$                                                                                             (I)

Here, $B_{net}$ is the net magnetic field, $B_{\text{left}}$ is the magnetic field due to left and $B_{\text{right}}$ is the magnetic field due to right.

Write the equation magnetic field due to left,

  $B_{\text{left}}=\frac{{\mu }_{0}I}{2\pi r_1}$                                                                                                        (II)

Here, ${\mu }_0$ is the permeability, $I$ is the current and $r_1$ is the distance.

Write the equation magnetic field due to right,

  $B_{\text{right}}=\frac{{\mu }_{0}I}{2\pi r_2}\cos \theta$                                                                                             (III)

Here, $r_2$ is the distance and $\theta$ is the angle.

Write the equation of the distance,

  $\begin{array}{c}{r}_2=\sqrt{d^2+{\left(d/2\right)}^2}\\ =\left(d/2\right)\sqrt{5}\end{array}$                                                                                            (IV)

Here, $d$ is the side of the square.

Write the equation of the distance,

  $r_1=\left(d/2\right)$                                                                                                         (V)

Write the equation of the angle.

  $\begin{array}{c}\theta ={\tan }^{-1}\left(d/\left(d/2\right)\right)\\ ={\tan }^{-1}\left(2\right)\end{array}$                                                                                         (VI)

Rewrite the expression from equation (I)

  $B_{net}=2\left(\frac{{\mu }_{0}I}{2\pi \left(\left(d/2\right)\right)}\right)+2\left(\frac{{\mu }_{0}I}{2\pi \left(\left(d/2\right)\sqrt{5}\right)}\cos \left({\tan }^{-1}\left(2\right)\right)\right)$                            (VII)

Conclusion:

Substitute $4\pi \times {10}^{-7}\text{T-m/A}$ for ${\mu }_0$, $10.0\text{A}$ for $I$ and $0.10\text{m}$ for $d$ in equation (VII) to find $B_{net}$.

  $\begin{array}{c}{B}_{net}=2\left(\frac{\left(4\pi \times {10}^{-7}\text{T-m/A}\right)\left(10.0\text{A}\right)}{2\pi \left(\left(0.10\text{m}\right)/2\right)}\right)+2\left(\frac{\left(4\pi \times {10}^{-7}\text{T-m/A}\right)\left(10.0\text{A}\right)}{2\pi \left(\left(\left(0.10\text{m}\right)/2\right)\sqrt{5}\right)}\cos \left({\tan }^{-1}\left(2\right)\right)\right)\\ =2\left(\frac{\left(4\pi \times {10}^{-7}\text{T-m/A}\right)\left(10.0\text{A}\right)}{2\pi }\right)\left[\left(\frac{1}{\left(\left(0.10\text{m}\right)/2\right)}\right)+\left(\frac{\cos \left(63.4°\right)}{\left(\left(\left(0.10\text{m}\right)/2\right)\sqrt{5}\right)}\right)\right]\\ =96\text{μT}\end{array}$

The direction of the magnetic field is directed to the right.

Thus, the magnetic field at point R is $\underset{\_}{96\text{μT and directed to the right}}$.