Question

In an experiment designed to measure the Earth's magnetic field using the Hall effect, a copper bar 0.410 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 4.75 10-12 V, what is the magnitude of the Earth's magnetic field at this location?

Answer 1

`B=3.2* 10^(-5)\ T`

Step-by-step explanation:

Given that

t = 0.410 cm

`n = 8.46 * 10^(28)\ electrons/m^3`

I= 8 a

`V_h= 4.75* 10^(-12)\ V`

`q=1.6* 10^(-19)\ C`

We know that magnetic field given as

`B=(nqtV_h)/(I)`

`B=(8.46 *  10^(28)* 1.6* 10^(-19)* 0.41* 10^(-2)* 4.75* 10^(-12))/(8)`

`B=3.2* 10^(-5)\ T`