If a current of 8.00 A in the conductor results in a Hall voltage of 5.85 10-12 V, what is the magnitude of the Earth's magnetic field at this location?

+2 votes

In an experiment designed to measure the Earth's magnetic field using the Hall effect, a copper bar 0.480 cm thick is positioned along an east–west direction. Assume n = 8.46 × 1028 electrons/m3 and the plane of the bar is rotated to be perpendicular to the direction of

Bimage.

If a current of 8.00 A in the conductor results in a Hall voltage of 5.85 image 10-12 V, what is the magnitude of the Earth's magnetic field at this location?
µT  

asked Feb 19, 2012 in Physics by apush ~Expert~ (552 points)
    

1 Answer

+2 votes
 

Hall effect voltage is given by the following equation:

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 get:

Plugging in the values we get:

 

B = [(8.46 × 1028 e/m3)(1.60 × 10−19 C/e) (0.00480 m)(5.85 image 10-12 V)] / (8 A)

= 4.75 *10^-5 T  = 47.5 µT

 

answered Feb 24, 2012 by pokemonmaster ~Expert~ (3,856 points)

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