Chapter 28, Problem 8P

At the location of the compass, the magnetic field caused by the wire will point to the west, and the Earth’s magnetic field points due North. The compass needle will point in the direction of the NET magnetic field.

$\begin{array}{l}{B}_{\text{wire}}=\frac{{\mu }_{0}I}{2\pi r}=\frac{\left(4\pi \times {10}^{-7}\text{T}\cdot \text{m/A}\right)(43\text{A})}{2\pi (0.18\text{m})}=4.78\times {10}^{-5}\text{T}\\ \theta ={\tan }^{-1}\frac{B_{\text{Earth}}}{B_{\text{wire}}}={\tan }^{-1}\frac{4.5\times {10}^{-5}\text{T}}{4.78\times {10}^{-5}\text{T}}=\boxed{43°\text{N}\text{of}\text{W}}\end{array}$

 (II) A horizontal compass is placed 18 cm due south from a straight vertical wire carrying a 43-A current downward. In what direction docs the compass needle point at this location? Assume the horizontal component of the Earth’s field at this point is 0.45 × 10^-4 T and the magnetic declination is 0°.