In an experiment designed to measure the Earth’s magnetic...

In an experiment designed to measure the Earth’s magnetic field using the Hall effect, a copper bar 0.500 cm thick is positioned along an east–west direction. Assume $n=8.46 \times 10^{28}$ electrons $/ \mathrm{m}^{3}$ and the plane of the bar is rotated to be perpendicular to the direction of $\overrightarrow{\mathbf{B}}$ . If a current of 8.00 $\mathrm{A}$ in the conductor results in a Hall voltage of $5.10 \times 10^{-12} \mathrm{V},$ what is the magnitude of the Earth's magnetic field at this location?

Key Concepts

Conductor Geometry

Conductor geometry, especially dimensions like thickness, plays a crucial role in the manifestation of the Hall effect. The thickness of the conductor directly influences the separation distance over which the Hall voltage develops, and it is an important factor in the mathematical relationship used to relate the Hall voltage to the magnetic field strength.

Charge Carrier Density

Charge carrier density refers to the number of charge carriers (such as electrons) per unit volume in a material. It is an essential parameter in determining the electrical and magnetic behavior of conductors, particularly in the context of the Hall effect where it influences the magnitude of the Hall voltage for a given current and magnetic field.

Magnetic Field Measurement

Using the Hall effect to measure magnetic fields involves relating the induced Hall voltage to the magnetic field strength, current, charge carrier density, and geometric dimensions of the conductor. This method provides a practical, non-intrusive means to determine magnetic fields in various applications, including environmental and geophysical studies.

Hall Effect

The Hall effect describes the generation of a voltage difference (the Hall voltage) across an electrical conductor when a magnetic field is applied perpendicular to the flow of current. This effect arises because the magnetic field exerts a Lorentz force on the moving charge carriers, deflecting them to one side of the conductor and creating a measurable potential difference. It is widely used in sensors and measurement devices to determine magnetic field strengths.

Lorentz Force

The Lorentz force is the force experienced by a moving charge within a magnetic field, which acts perpendicular to both the velocity of the charge and the magnetic field. This force is the fundamental mechanism behind the Hall effect, as it causes the deflection of charge carriers in a conductor, leading to charge accumulation and the subsequent creation of the Hall voltage.

Transcript

0:00 All righty.

00:01 So this problem consists of a pretty cool experiment that's implemented to measure the earth's magnetic field using this very cool thing called the hall effect.

00:11 And there's a copper bar that's about a half centimeter thick and it's positioned along the east -west direction.

00:16 And they give you the number of electrons per meter cubed, which i can write here as a little n, little n, and that equals 8 .46 times 10, 2d28.

00:31 Electrons per meter cubed, that e minus being the electrons.

00:36 And the plane of the bar is rotated to be perpendicular to the direction of the b field.

00:41 All right, good we know.

00:42 And we also know that there's a current 8 amps in the conductor that results in a hull voltage of 5 .1.

00:49 So i want to say my voltage of 5 .10 times 10 to minus 12 volts.

00:58 All right and of course i remember we also have the thickness of the copper bar that's a half centimeter we'll implement that when we need to on the we also have the current i equals eight all right and bans i should say a professional here cool all right so let's go ahead we know that the magnitude of the earth's magnetic field is found by solving this guy right here we've seen us before here one actually gives little space here to work these guys a little bit separated there cool and that's this haul voltage you could say your haul voltage which you could call this little be professional delta b hall which is equal to the r sub h which is the hall coefficient multiplied by i b and divided by the thickness t headset thickness half centimeter i talked about earlier and we rewrite this as i times b divided by n q times t and the way i did that is i replaced that hull voltage or assuming the hall coefficient with one over nq totally okay totally correct let's go ahead now keep moving and uh go ahead and get to a solution so what we got to do is take this equation right here and solve it for b we're solving this for b actually i'm sorry be more exact here these guys solving these guys here all right or that freaking b there here all right so once we do that get this delta v all voltage n q and t was all come across the nqt moving that this side and then we're going to have the i that's going to move to the denominator over here and all we got to do is plug in what we know 5 .1 times 10 to the minus 12 volts 8 .46 times 10 2d28 meters minus per meter cubed 1 .6 times 10 to minus 19 coulomes...

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