Question

In an experiment designed to measure the Earth’s magnetic field using the Hall effect, a copper bar 0.500 cm thick is positioned along an east–west direction. Assume n = 8.46 × 10 28 electrons/m 3 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 5.10×10 -12 V, what is the magnitude of the Earth’s magnetic field at this location?

Answer 1

Answer:

$: \implies \sf{ \red{B= 4.197\times10^{-5}T}}$

Explanation:

We know that

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

Hall effect voltage and magnetic field are defined as,

$\sf{\Delta V_H = \dfrac{IB}{nqt}}$

$\sf \blue{B=\dfrac{nqt\Delta V_h}{I}} \: { \bigstar}$

So,

$\implies \sf{B=\dfrac{(8.48\times10^{28})(1.6\times10^{-19})(0.0055)(4.5\times10^{-12})}{8}}$

$\implies \sf{B= 0.00004197}$

$\implies \sf{ \red{B= 4.197\times10^{-5}T}}$

Answer 2

Answer:

$: \implies \sf{ \red{B= 4.197\times10^{-5}T}}$

Explanation:

We know that

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

Hall effect voltage and magnetic field are defined as,

$\sf{\Delta V_H = \dfrac{IB}{nqt}}$

$\sf \blue{B=\dfrac{nqt\Delta V_h}{I}} \: { \bigstar}$

So,

$\implies \sf{B=\dfrac{(8.48\times10^{28})(1.6\times10^{-19})(0.0055)(4.5\times10^{-12})}{8}}$

$\implies \sf{B= 0.00004197}$

$\implies \sf{ \red{B= 4.197\times10^{-5}T}}$