Chapter 20, Problem 46P

To determine

The magnitude and direction of the magnetic field at the midpoint of the side of the triangle between wire M and wire N.

Answer to Problem 46P

The net magnetic field at the midpoint of the side of the triangle between wire M and wire N is $1.75\times {10}^{-4} \text{T}$ pointing $14°$ below the horizontal.

Explanation of Solution

Given:

Three wires are at $3.8 \text{cm}$ away from each other with the current of 8 A. The direction of current in wire M is opposite to that in the wires N and P.

  

Formula used:

The magnetic field in the wire is calculated as

  $B=\frac{{\mu }_{0}I}{2\pi r}$

Calculation:

Consider the figure shown below.

  

There will be three magnetic fields to sum- one from each wire. Each magnetic field points perpendicularly to the line connecting the wires to the midpoint. The two fields due to M and N are drawn slightly separated from each other but should be co-linear. Thus, the net magnetic field is the vector sum of three individual ones:

  $\overrightarrow{B_{net}}=\overrightarrow{B_M}+\overrightarrow{B_N}+\overrightarrow{B_P}$

Now, the individual magnetic field across each wire is:

  $\begin{array}{l}{B}_M=B_N=\frac{{\mu }_{0}I}{2\pi r_M}\\ B_M=B_N=\frac{{\text{(4π×10}}^{\text{-7}}\text{ T-m/A)}}{\text{2π}}\frac{\text{8 A}}{\text{(0}\text{.019 m)}}\\ B_M=B_N=8.421\times {10}^{-5}\text{ T}\end{array}$

  $\begin{array}{l}{B}_P=B=\frac{{\mu }_{0}I}{2\pi r_P}=\frac{{\text{(4π×10}}^{\text{-7}}\text{ T-m/A)}}{\text{2π}}\frac{\text{8 A}}{\sqrt{\text{3}}\text{(0}\text{.019 m)}}\\ B_P=4.862\times {10}^{-5} \text{T}\end{array}$

Since individual magnetic fields are calculated. Now, the net horizontal and vertical component of magnetic fields is:

  $\begin{array}{l}{B}_{net x}=2B_M\cos 30°+B_P\cos 60°\\ B_{net x}=2(8.421\times {10}^{-5})\cos 30°+(4.862\times {10}^{-5})\cos 60°\\ B_{net x}=1.702\times {10}^{-4} \text{T}\end{array}$

  $\begin{array}{l}{B}_{net y}=-2B_M\sin 30°+B_P\sin 60°\\ B_{net y}=-2(8.421\times {10}^{-5})\sin 30°+(4.862\times {10}^{-5})\sin {60}^{\circ }\\ B_{net y}=-4.21\times {10}^{-5} \text{T}\end{array}$

Now, the net magnetic field is:

  $\begin{array}{l}{B}_{net}=\sqrt{B_{net x}^2+B_{net y}^2}=\sqrt{{(1.702\times {10}^{-4})}^2+{(-4.21\times {10}^{-5})}^2}\\ B_{net}=1.75\times {10}^{-4} \text{T}\end{array}$ ; $\begin{array}{l}{\theta }_{net}={\tan }^{-1}\frac{B_{net y}}{B_{net x}}={\tan }^{-1}\frac{\text{-4}{\text{.21×10}}^{\text{-5}}\text{T}}{\text{1}{\text{.702×10}}^{\text{-4}}\text{T}}\\ {\theta }_{net}=-14°\end{array}$

Conclusion:

The net magnetic field at the midpoint of the side of the triangle between wire M and wire N is $1.75\times {10}^{-4} \text{T}$ pointing $14°$ below the horizontal.