A rectangular copper strip 1.5 cm wide and 0.10 cm thick carries a current of 5.

Question

A rectangular copper strip 1.5 cm wide and 0.10 cm thick carries a current of 5.0 A. Find the Hall voltage for a 1.2-T magnetic field applied in a direction perpendicular to the strip.Assuming that one electron per atom is available for conduction, we can take the charge carrier density to be 8.46  1028electrons/m3. Supposing the copper strip is 14 cm long, we can also measure the ohmic voltage drop across the strip along the direction of the current flow. This potential difference is typically much larger than the Hall voltage. What value of B will make the Hall voltage equal to 10% of the voltage drop along the length of the copper strip?

Explanation / Answer

I= neAv

==> v=I/neA=5/(8.46 E28*1.6 E-19*1.5*0.1 E-4) =2.46 E-5 m/s

now evB=e*E

==> v*B=Voltage/d

==>Halls voltage= d*v*B=1.2*0.1 E-3*2.46 E-5=0.2955 E-8 Volts

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