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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?

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Answer:

[tex]B= 4.197*10^-5T[/tex]

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,

[tex]\Delta V_H = \frac{IB}{nqt}[/tex]

[tex]B=\frac{nqt\Delta V_h}{I}[/tex]

So,

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

[tex]B= 0.00004197[/tex]

[tex]B= 4.197*10^{-5}T[/tex]

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