The two springs in FIGURE P29.72 each have a spring constant of 10 N/m. They are compressed by 1.0 cm when a current passes through the wire. How big is the current?

The diagram is as follows:

Given that the spring constant of each spring is $10\text{N/m}$, the springs are compressed by 1.0 cm, the value for the magnetic field is 0.5 T and the width within which the magnetic field is active is 20 cm..

The formula to calculate the force applied on the rod due to the compression of the springs is given by

$F=2kx.............................\left(1\right)$

Here, $F$is the spring force, $k$is the spring constant and $x$is the displacement of each spring.

The formula to calculate the magnetic force on the rod is given by

$F_B=ilB................................\left(2\right)$

Here, $F_B$is the magnetic force, $i$is the current through the rod, $l$is the length of the rod which is within the magnetic field and $B$isthe magnetic field.

At equilibrium, both the spring force and the magnetic force on the rod balance each other.

Equate equation (1) and equation (2) and simplify to obtain the required current.

$\begin{array}{rcl}2kx& =& ilB\\ i& =& \frac{2kx}{lB}.........................\left(3\right)\end{array}$

Substitute $10\text{N/m}$for $k$, $1.0\text{cm}$for $x$, $0.5\text{T}$for $B$and $20\text{cm}$for $l$into equation (3) to calculate the required current.

$\begin{array}{rcl}i& =& \frac{2\times 10\text{N/m}\times 1.0\text{cm}\times {\displaystyle \frac{1\text{m}}{100\text{cm}}}}{20\text{cm}\times \frac{1\text{m}}{100\text{cm}}\times 0.5\text{T}}\\ & =& 2.0\text{A}\end{array}$

The required current is$2.0\text{A}$.