A thin rectangular strip of semiconductor has a width of five centimetres and cross-sectional area of two millimetres squared. The strip is connected to a battery and placed in a magnetic field perpendicular to the surface of the strip. A Hall voltage of 12.5 volts is produced across the strip, associated with a drift velocity of 50 metres per second. What is the magnitude of magnetic field applied to the strip?
We can call the width of the strip 5 centimetres 𝑤. We’ll call the Hall voltage produced across the strip 12.5 volts 𝑣 sub 𝐻. And the drift velocity of 50 metres per second we’ll call 𝑣 sub 𝑑. We want to know the magnitude of the magnetic field applied to the strip. We’ll call that 𝐵. Let’s start by drawing a sketch of this rectangular strip.
Looking at a cross section of our semiconductor strip, which has a width of 𝑤, and has a potential difference from one side of the other of 𝑣 sub 𝐻, our Hall voltage, we see that there is a magnetic field which acts perpendicularly to the strip, which helps to direct charges along the strip as they come out — coming out of the page from our perspective.
The Hall voltage 𝑣 sub 𝐻 is given by an equation that relates some of the terms in our scenario. It’s equal to the magnetic field 𝐵 times the drift velocity 𝑣 sub 𝑑 multiplied by the width of the strip 𝑤. Since we want to solve for 𝐵 the magnetic field, we can rearrange this equation and find 𝐵 is equal to the hall voltage divided by the drift velocity 𝑣 sub 𝑑 multiplied by the width of the strip 𝑤.
Since we’re given each of these three values in our statement, we can plug in and solve for 𝐵 now. When we do, being careful to use units of metres for our width 𝑤, and enter these values on our calculator, we find that to one significant figure 𝐵 is five tesla. That’s the strength of the magnetic field acting perpendicular to the semiconductor strip.