This is College Physics Answers with Shaun Dychko. The EMF of a Hall Effect probe will be the magnetic field strength times the width of the probe times the speed of the charge carriers through it and these things are not going to change when the probe is put in the different magnetic field. So they don't need subscripts when we talk about the Hall Effect voltage in each of the two cases. So in the first case, it's in the first metallic field times <i>l</i> times <i>v</i>. And then on the second case, we have a different voltage and a different magnetic field <i>B2</i> times <i>l</i> times <i>v</i> and we'll just say that <i>lv</i> is <i>E1</i> over <i>B1</i>. We'll solve for it here by dividing both sides by magnetic field strength one. And then we'll substitute for <i>lv</i> in EMF two. And so EMF two is going to be the magnetic field strength in case two times EMF one divided by magnetic field strength in case one. So that is 0.15 Tesla times one times ten to the minus six voltage divided by two Tesla. And this gives 7.5 times ten to the minus eight Volts will be the EMF in the 0.15 Tesla magnetic field.