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

Answer:

$B=3.2\times 10^{-5}\ T$

Explanation:

Given that

t = 0.410 cm

$n = 8.46 \times 10^{28}\ electrons/m^3$

I= 8 a

$V_h= 4.75\times 10^{-12}\ V$

$q=1.6\times 10^{-19}\ C$

We know that magnetic field given as

$B=\dfrac{nqtV_h}{I}$

$B=\dfrac{8.46 \times 10^{28}\times 1.6\times 10^{-19}\times 0.41\times 10^{-2}\times 4.75\times 10^{-12}}{8}$

$B=3.2\times 10^{-5}\ T$