Chapter 20, Problem 35P

To determine

The net magnetic field of the wire.

Answer to Problem 35P

The net magnetic field of the wire is $4.1\times {10}^{-5} \text{T}$ pointing $11°$ below the horizontal.

Explanation of Solution

Given:

Current carried by wire is 22 A in the North direction.

Earth’s magnetic field is $5\times {10}^{-5} \text{T}$ points in North but downward $37°$ below the horizontal.

Formula used:

Magnetic field in the current-carrying wire is calculated as

  $B=\frac{{\mu }_{0}I}{2\pi r}$

Calculation:

Consider the diagram shown below.

  

As shown in the diagram, magnetic field of wire is facing straight upward of magnitude.

  $\begin{array}{l}{B}_{wire}=\frac{{\mu }_{0}I}{2\pi r}\\ B_{wire}=\frac{{\text{(4π×10}}^{\text{-7}}\text{ T}\cdot \text{m/A)(22 A)}}{\text{2π(0}\text{.2 m)}}\\ B_{wire}=2.2\times {10}^{-5} \text{T}\end{array}$

Earth’s horizontal component of magnetic field gives the net horizontal component of magnetic field that is,

  $\begin{array}{c}{B}_{net x}=B_{Earth}\cos 37°\\ =(5\times {10}^{-5})\cos 37°\\ B_{net x}=4\times {10}^{-5} \text{T}\end{array}$

Now, the net vertical component of magnetic field is

  $\begin{array}{l}{B}_{net y}=B_{wire}-B_{Earth}\sin 37°\\ B_{net y}=-0.81\times {10}^{-5} \text{T}\end{array}$

The net magnetic field is the vector sum of both horizontal and vertical components. Thus,

  $\begin{array}{l}{B}_{net}=\sqrt{B_{net x}^2+B_{net y}^2}\\ B_{net}=\sqrt{{(4\times {10}^{-5})}^2+{(-0.81\times {10}^{-5})}^2}\\ B_{net}=4.1\times {10}^{-5} \text{T}\end{array}$

Now, the direction of the net magnetic field of the wire is

  $\begin{array}{c}\theta ={\tan }^{-1}\frac{B_{net y}}{B_{net x}}\\ ={\tan }^{-1}\frac{(-0.81\times {10}^{-5})}{(4\times {10}^{-5})}\\ \theta =11°\end{array}$

Conclusion:

Therefore, the net magnetic field of the wire is $4.1\times {10}^{-5} \text{T}$ pointing $11°$ below the horizontal.