Question

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?

Answer 1

`B= 4.197*10^-5T`

Step-by-step 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 = (IB)/(nqt)`

`B=(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`