Chapter 20, Problem 50P

To determine

The force on the wire assuming the solenoid’s field points due east.

Answer to Problem 50P

The force on the wire is $0.10\text{ N}$ .

Explanation of Solution

Given:

A solenoid is $15\text{ cm}$ with $550$ turns carrying $33\text{ A}$ current. A wire of $3\text{ cm}$ carrying $22\text{ A}$ current downwards cuts through the center of the solenoid along the diameter.

Formula used:

The magnetic force is calculated as

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

Calculation:

The magnetic force on the wire will be perpendicular to the direction of current that is

  $\begin{array}{l}{\text{F=I}}_{\text{wire}}{\text{l}}_{\text{wire}}{\text{B}}_{\text{solenoid}}={\text{I}}_{\text{wire}}{\text{l}}_{\text{wire}}\frac{{\mu }_0{\text{I}}_{\text{solenoid}}\text{N}}{{\text{l}}_{\text{solenoid}}}\\ \text{F=(22 A)(0}\text{.030 m)}\frac{(4\pi \times {10}^{-7}\text{ T}·\text{m/A})(33\text{ A})(550)}{(0.15\text{ m})}\\ \text{F=0}\text{.10 N}\end{array}$

Conclusion:

Thus, the force experienced on the wire is $0.10\text{ N}$ .