A Hall probe calibrated to read $\mathbf{1}\mathbf{.}\mathbf{00}\mathbf{}\mathbf{\mu V}$a when placed in a $\mathbf{2}\mathbf{.}\mathbf{00}\mathbf{‐}\mathbf{T}$field is placed in a $0.150‐\mathrm{T}$ field. What is its output voltage?

The induced voltage in a Hall probe is ${\mathrm{E}}_1=1.00\mathrm{\mu V}$

The magnetic field strength which can induce ${\mathrm{E}}_1$voltage is ${\mathrm{B}}_1=2.00\mathrm{T}$

Test magnetic field where we need to find the induced voltage is${\mathrm{B}}_2=0.150\mathrm{T}$

The Hall voltage generated across an electrical conductor that is transverse to the current and perpendicular to the applied magnetic field can be given as,

$\mathbf{E}\mathbf{}\mathbf{=}\mathbf{BLv}\mathbf{}\mathbf{}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{}\left(1\right)$

Where, B, is the applied magnetic field in the perpendicular direction, L and v is the length and speed of the moving conductor.

If we assume that the length of the conductor is L and velocity is v then for case 1, the induced voltage can be expressed by using equation (1),

${\mathrm{E}}_1={\mathrm{B}}_1\mathrm{Lv}$……………..(2)

Similarly, for case 2, we can get the induced voltage sing equation (1) such that,

${\mathrm{E}}_2={\mathrm{B}}_2\mathrm{Lv}$…………….(3)

Now, taking the ratio of equations (2) and (3) will result in,

$$\begin{gathered} \frac{{\mathrm{E}}_2}{{\mathrm{E}}_1}=\frac{{\mathrm{B}}_2}{{\mathrm{B}}_1} \\ {\mathrm{E}}_2=\frac{{\mathrm{B}}_2}{{\mathrm{B}}_1}\times {\mathrm{E}}_1 \\ {\mathrm{E}}_2=\frac{0.150\mathrm{T}}{2.00\mathrm{T}}\times 1.00\mathrm{\mu V} \\ {E}_2=0.075\mu V \end{gathered}$$

Therefore, the induced hall voltage in field B₂ is $0.075\mathrm{\mu V}$ .