Chapter 20, Problem 40P

To determine

The magnitude and direction of the current in the wire.

Answer to Problem 40P

The current in the wire is $19.4\text{ A}$ facing down.

Explanation of Solution

Given:

A compass needle changes its direction from $23°\text{ E of N}$ to $55°\text{E of N}$ from the vertical wire, when placed $12\text{ cm}$ east of wire.

Earth’s magnetic field is $0.5\times {10}^{-4}\text{ T}$ horizontally.

Formula used:

According to the law of sines,

  $\frac{\overrightarrow{{\text{F}}_{\text{A}}}}{\sin {\theta }_{\text{A}}}=\frac{\overrightarrow{{\text{F}}_{\text{B}}}}{\sin {\theta }_{\text{B}}}$

Calculation:

Considering, both the changes in the magnetic field around the wire is in the Earth’s magnetic field. According to the diagram below, compass needle is moving towards south direction. Thus, the net magnetic field due to wire must be towards south direction.

  

Using the geometry of the figure, angles are $\theta =23°$ , $\theta +\varphi =55°$ , and $\beta +\theta +\varphi =180°$ .

So, $\varphi =32°$ , and $\beta =125°$ . Now, using the law of sines, magnetic field of the wire will be

  $\begin{array}{l}\frac{{\text{B}}_{\text{wire}}}{\sin \varphi }=\frac{{\text{B}}_{\text{Earth}}}{\sin \beta }\Rightarrow {\text{B}}_{\text{wire}}={\text{B}}_{\text{Earth}}\frac{\sin \varphi }{\sin \beta }=\frac{{\mu }_0\text{I}}{2\pi \text{r}}\\ {\text{I = B}}_{\text{Earth}}\frac{\sin \varphi }{\sin \beta }\frac{2\pi \text{r}}{{\mu }_0}\\ \text{I = (5}\times {\text{10}}^{-5}\text{}\text{T)}\frac{\sin {32}^0}{\sin {125}^0}\frac{2\pi (0.120\text{ m})}{4\pi \times {\text{10}}^{-7}\text{}\text{T}\cdot \text{m/A}}\\ \text{I =}19.4\text{ A}\end{array}$

Since the magnetic field in the wire is facing towards the south direction. Thus, current in the wire will be towards down.

Conclusion:

Thus, the current in the wire is $19.4\text{ A}$ facing down.