Chapter 20, Problem 41P

To determine

The magnitude and the direction of the net force on the loop.

Answer to Problem 41P

The net force on the loop is $2.6\times {10}^{-6}\text{ N}$ towards the wire.

Explanation of Solution

Given:

The given circuit is shown below.

  

Formula used:

The magnetic force is calculated as

  $F=\frac{{\mu }_0{I}_1{I}_2}{2\pi d_2}{l}_2$

Calculation:

Only forces experienced on the nearest wire and farther wire of the loop concerning the straight wire. Using the right-hand rule, the magnetic force in the top wire of the loop will be towards upwards direction and in the far away wire of the loop, it will be towards downwards. Magnetic force experienced by the top wire will be stronger than it is experienced by the bottom wire due to increased distance. Thus, net magnetic force experienced on the loop is

  $F_{net}=F_{near}-F_{far}=\frac{{\mu }_0{I}_1{I}_2}{2\pi d_{near}}{l}_{near}-\frac{{\mu }_0{I}_1{I}_2}{2\pi d_{far}}{l}_{far}$

The length of the top and bottom wire is same that is $l_{near}=l_{far}=l$ . Thus,

  $\begin{array}{l}{F}_{net}=\frac{{\mu }_0{I}_1{I}_2}{2\pi }l\left(\frac{1}{d_{near}}-\frac{1}{d_{far}}\right)\\ F_{net}\text{=}\frac{4\pi \times {10}^{-7}\text{ T}\cdot \text{m/A}}{2\pi }{(}^{2.5}(0.100\text{ m})\left(\frac{1}{0.030\text{ m}}-\frac{1}{0.080\text{ m}}\right)\\ F_{net}\text{=}2.6\times {10}^{-6}\text{ N}\end{array}$

Since top wire has stronger magnetic forces comparatively lower wire. Thus, net magnetic force will be in the upwards direction that is towards the wire.

Conclusion:

The net force on the loop is $2.6\times {10}^{-6}\text{ N}$ towards the wire.