Tutorial Exercise In an experiment designed to measure the Earth\'s magnetic fie

Question

Tutorial Exercise In an experiment designed to measure the Earth's magnetic field using the Hall effect, a copper bar 0.405 cm thick is positioned along an east-west direction. Assume n 8.46 x 1028 electrons/m3 and the plane of the bar is rotated to be perpendicular to the direction of B. If a current of 8.00 A in the conductor results in a Hall voltage of 4.27 x 10- V, what is the magnitude of the Earth's magnetic field at this location? 12 Part 1 of 3 - Conceptualize From this Table of Approximate Magnetic Field Magnitudes, the magnitude of the Earth's magnetic field is about 50 ?T, we expect a result with this order of magnitude. Part 2 of 3 - Categorize We will find the magnitude of the magnetic field from the equation for the Hall effect voltage. Part 3 of 3 - Analyze Hall effect voltage is given by the following nqt where I is the current, n is the volume density of electrons, B is the magnitude of the magnetic field, and t is the thickness of the copper bar. Solving for the magnitude of the magnetic field, we have nqtAVH 8.46 x 1028 e-/m3 )(1.60 x 10-19 C/e × 10-12V 5 HT.

Explanation / Answer

Hall effect voltage is given by:

dVh = I*B/(nqt)

B = n*q*t*dVh/I

t = thickness = 0.405 cm = 0.00405 m

I = 8.00 A

Using given values:

B = (8.46*10^28 e/m^3)(1.6*10^-19 C/e)(0.00405 m)(4.27*10^-12 V)/(8.00 A)

B = 2.926*10^-5 T

B = 29.26 uT

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