Question

The figure below shows a portion of a silver ribbon with z₁ = 11.8 mm and y₁ = 0.23 mm, carrying a current of 102 A in the +x-direction. The ribbon lies in a uniform magnetic field, in the y-direction, with magnitude 1.09 T. Apply the simplified model of the Hall effect presented in Section 27.9. They are 5.85 10²⁸ free electron per cubic meter.

The figure below shows a portion of a silver ribbon with z1 11.8 mm and yi-0.23 mm, carrying a current of 102 A in the +x-direction. The ribbon lies in a uniform magnetic field, in the y-direction, with magnitude 1.09 T. Apply the simplified model of the Hall effect presented in Section 27.9 They are 5.85 x 1028 free electron per cubic meter. (a) Find the magnitude of the drift velocity of the electrons in the x-direction. m/s (b) Find magnitude and the direction of the electric field in the z-direction due to the Hall effect. magnitude N/C +x-direction x-direction +y-direction y-direction direction O +z-direction z-direction (c) Find the Hall emf.

Answer 1

The current density is, 102 A -37.58x10 A/m2 (i 1.8x10" m)(0.23x10-3 m) A The magnitude of the drift velocity of the electrons is, 37.58x106 4n 5.85x10 1.6x101) -4.01x10-3 m/s Therefore, the magnitude of the drift velocity is 4.01x10 m/s The direction of the drift velocity is along positive x-axis because the current is along positive X-aXis. The electric field is, = (4.01x10-3 m/s)xx(1.09 T) - 4.37 x103 N/C2 Therefore, the magnitude of the electric field is 4.37x10- N/C and the direction of the ele field is +z-direction The expression for the hall emf is, e E, (2) - (4.37 x103 (11.8x10-3 m - 51.57x10-6 v Therefore, the induced emf is 51.57x10-6 v