91Ó°ÊÓ

The given is the $1.9\mathrm{μ}V$voltage across the $55mT$magnetic field and $2.8\mathrm{μ}V$voltage across the different magnetic fields. The objective is to find the strength of the second field.

When a magnetic field is applied to the moving charges inside a conductor, the hall voltage is the steady-state potential difference between the two surfaces of the conductor. By using an equation of the form, the hall voltage is connected to the current inside the conductor and the magnetic field

$∆V_H=\frac{IB}{tne}$

Where $n$denotes charge density, $t$denotes the thickness of the conductor, $e$denotes charge of the electron, and $B$denotes the exerted magnetic field. As shown by the equation, the Hall voltage is directly proportional to the magnetic field

$∆V_H\mathrm{Î\pm }B$

When the current is the same for both conductors, we can get an equation for two instants between the Hall voltages and the magnetic field in the form

$$\begin{gathered} \frac{∆V_{1,H}}{∆V_{2,H}}=\frac{B_1}{{B_2}_{}} \\ {B}_2=\left(\frac{∆V_{2,H}}{∆V_{1,H}}\right)B_1 \end{gathered}$$

If we have $∆V_{1,H}=1.9\mathrm{μ}V$at $B_1=55mT$, we can get $B_2$at $∆V_{2,H}=2.8\mathrm{μ}V$by

$$\begin{gathered} B_2=\left(\frac{∆V_{2,H}}{∆V_{1,H}}\right)B_1 \\ =\left(\frac{2.8\mathrm{μ}V}{1.9\mathrm{μ}V}\right)\left(55mT\right) \\ =81mT \end{gathered}$$