A rectangular copper strip 1.5cm wide and 0.10cn thick carries a current of 5.0A. Find the Hall voltage for a 1.2T magnetic field applied in a direction perpendicular to the loop

4.4345× 10^-7V

Explanation:

The computation of the half voltage for a 1.2T magnetic field applied is shown below

The volume of one mole of copper is

v = m ÷p

= 63.5 ÷ 8.92

= 7.12cm

Now the density of free electrons in copper is

n = Na ÷ V

= 6.02 × 10^23 ÷ 7.12

= 8.456× 10^28/m^3

Now the half voltage is

= IB ÷ nqt

= (5 × 1.20) ÷ (8.456× 10^28 × 1.6 × 10^-19 × 0.1× 10^-2)

= 4.4345× 10^-7V

probably not the "magnets"in the magnet, so to speak, are thought to be electrons orbiting in shells. vurrent loops two sides to the loop

your answer is c! hope this : )