Question

Magnetic fields are sometimes measured by balancing magnetic forces against known mechanical forces. Your task is to measure the strength of a horizontal magnetic field using a 12-cm-long rigid metal rod that hangs from two nonmagnetic springs, one at each end, with spring constants 1.3 N/m. You first position the rod to be level and perpendicular to the field, whose direction you determined with a compass. You then connect the ends of the rod to wires that run parallel to the field and thus experience no forces. Finally, you measure the downward deflection of the rod, stretching the springs, as you pass current through it. Your data are as follows: Current (A) Deflection (mm) 1.0 4 2.0 9 3.0 12 4.0 15 5.0 21 Use an appropriate graph of the data to determine the magnetic field strength.

Short answer

The magnetic field strength is$87mT$.

Step-by-step solution

Given information

Data given is shown below

Current (A)	Deflection (mm)
1.0	4
2.0	9
3.0	12
4.0	15
5.0	21

Explanation

When the current is on, the magnetic force $F_B=ILB$will be directed down by the right-hand rule. This will stretch the springs, which will produce an upward force equal in magnitude to the magnetic force.

There are two springs, so the upward force due to the stretched springs will be $F_{sp}=2k\left(∆y\right)$The rod is in equilibrium, so

$$\begin{gathered} 2k\left(∆y\right)=ILB \\ ∆y=\frac{LB}{2k} \end{gathered}$$

The graph of $∆y$vs $I$would produce a straight line whose slope is $\frac{LB}{2k}$and whose intercept is zero.

Magnetic field strength

The graph of $∆y$vs $I$is shown below.

It can be observed from the spreadsheet that the linear fit is reasonable and that the slope is $0.0040m/A$ and the intercept is very small.

$$\begin{gathered} B=\frac{slope\times 2k}{L} \\ =\frac{0.0040\times 2\times 1.3}{0.12} \\ =87mT \end{gathered}$$

Therefore, the magnetic field strength is$87mT$.