Chapter 20, Problem 45P

To determine

The magnetic force per unit length on each wire due to the other two wires.

Answer to Problem 45P

The magnetic force per unit length on wire M due to the other two wires is $5.8\times {10}^{-4}\text{ N/m at }{90}^0$ , on wire N due to the other two wires is $3.4\times {10}^{-4}\text{ N/m at }{300}^0$ , and on P due to the other two wires is $3.4\times {10}^{-4}\text{ N/m at }{240}^0$

Explanation of Solution

Given:

Consider the figure shown below.

  

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

Formula used:

The magnetic force in the wire is calculated as

  $\text{F}=\frac{{\mu }_0{\text{I}}_1{\text{I}}_2}{2\pi \text{r}}\text{a}$

Calculation:

Since direction of the current flowing in the wire M is opposite to the current flowing in the other two wires then magnetic force due to wire M will be repelling against the magnetic force induced by the other two wires. As shown in the diagram below, the horizontal component will cancel out since the horizontal component of the magnetic force applied on the other two wires will be equal and opposite. But, the vertical component will be the vector sum of both.

  

  $\begin{array}{c}\frac{F_{Mnety}}{l_M}=\frac{F_{MN}}{l_M}\cos {30}^0+\frac{F_{MP}}{l_M}\cos {30}^0\\ =\frac{{\mu }_0{I}_M{I}_p}{2\pi d_{MP}}(1m)\cos {30}^0+\frac{{\mu }_0{I}_M{I}_p}{2\pi d_{MP}}(1m)\cos {30}^0\\ =2\frac{(4\pi \times {10}^{-7}\text{T}\cdot \text{m/A})}{2\pi }\frac{{(8\text{ A})}^2}{(0.038\text{ m})}\cos {30}^{\circ }\\ =5.8\times {10}^{-4}\text{ N/m at }{90}^0\end{array}$

The force on wire N will be attractive towards wire P and a repelling away from wire M. The net force per unit length will be the vector sum of the horizontal and vertical force as shown in the diagram below.

  

  $\begin{array}{c}{F}_{Nnety}=-F_{NM}\cos {30}^0\\ =-\frac{{\mu }_0{I}_M{I}_N}{2\pi d_{MN}}\cos {30}^0\\ =-\frac{(4\pi \times {10}^{-7}\text{ T}\cdot \text{m/A})}{2\pi }\frac{{(8\text{ A})}^2}{(0.038\text{ m})}\cos {30}^0\\ =-2.917\times {10}^{-4}\text{ N/m}\end{array}$

  $\begin{array}{c}{F}_{Nnetx}=-F_{NM}\sin {30}^0+F_{NP}\\ =-\frac{(4\pi \times {10}^{-7}\text{ T}\cdot \text{m/A})}{2\pi }\frac{{(8\text{ A})}^2}{(0.038\text{ m})}\sin {30}^0+\frac{(4\pi \times {10}^{-7}\text{ T}\cdot \text{m/A})}{2\pi }\frac{{(8\text{ A})}^2}{(0.038\text{ m})}\\ =1.684\times {10}^{-5}\text{ N/m}\end{array}$

Now, the net magnetic force on N due to the other two wires is

  $\begin{array}{c}{F}_{Nnet}=\sqrt{F_{Nnetx}^2+F_{Nnety}^2}\\ =3.4\times {10}^{-4}\text{ N/m}\end{array}$

  $\begin{array}{l}\theta ={\tan }^{-1}\frac{F_{Nnety}}{F_{Nnetx}}={\tan }^{-1}\frac{-2.917}{1.684}\\ \theta ={300}^0\end{array}$

The force on wire P will be attractive towards wire N and repelling away from wire M. The net force per unit length will be the vector sum of the horizontal and vertical force as shown in the diagram below.

  

As shown in the above 2 figures, this is a mirror image of the previous solution. Thus, the net force is the same as $F_{P net}=3.4\times {10}^{-4} \text{N/m}$ and $\theta =240°$

Conclusion:

The magnetic force per unit length on wire M due to the other two wires is $5.8\times {10}^{-4}\text{ N/m at }{90}^0$ , on wire N due to the other two wires is $3.4\times {10}^{-4}\text{ N/m at }{300}^0$ , and on P due to the other two wires is $3.4\times {10}^{-4}\text{ N/m at }{240}^0$