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swat32
3 years ago
6

In an experiment designed to measure the Earth's magnetic field using the Hall effect, a copper bar 0.550 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 B. If a current of 8.00 A in the conductor results in a Hall voltage of 4.50 10-12 V, what is the magnitude of the Earth's magnetic field at this location?
Physics
1 answer:
motikmotik3 years ago
7 0

Answer:

B= 4.197*10^-5T

Explanation:

We know that the magnetic Field of the Earth is 50\mu T, so the equation for

Hall effect voltage und magnetic field are defined as,

\Delta V_H = \frac{IB}{nqt}

B=\frac{nqt\Delta V_h}{I}

So,

B=\frac{(8.48*10^{28})(1.6*10^{-19})(0.0055)(4.5*10^-{12})}{8}

B= 0.00004197

B= 4.197*10^{-5}T

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