The figure (Figure 1) shows a portion of a silver ribbon with z 1 = 13.49mm and

Question

The figure (Figure 1) shows a portion of a silver ribbon with z1 = 13.49mm and y1 = 1.36mm , carrying a current of 60A in the +x-direction. The ribbon lies in a uniform magnetic field, in the y-direction, with magnitude 2.2T . Apply the simplified model of the Hall effect.

Part A

If there are 5.811028 free electrons per cubic meter, find the magnitude of the drift velocity of the electrons in the x-direction.

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Part B

Find the magnitude of the electric field in the z-direction due to the Hall effect.

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Part D

Find the Hall emf.

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v_d = m/s T The figure (Figure 1) shows a portion of a silver ribbon with z1 = 13.49mm and y1 = 1.36mm , carrying a current of 60A in the +x-direction. The ribbon lies in a uniform magnetic field, in the y-direction, with magnitude 2.2T . Apply the simplified model of the Hall effect. Part A If there are 5.8110^8 free electrons per cubic meter, find the magnitude of the drift velocity of the electrons in the x-direction. SubmitMy AnswersGive Up Part B Find the magnitude of the electric field in the z-direction due to the Hall effect. SubmitMy AnswersGive Up Part D Find the Hall emf. SubmitMy AnswersGive Up

Explanation / Answer

current inx-direction I = 60 A

uniform magnetic field, in the y-direction B = 2.2 T

and y1 = 1.36 mm = 1.36*10-3 m

       z1 = 13.49 mm = 13.49*10-3 m

charge of the electron q = 1.6*10-19 C

a)

free electron density n = 5.85*10^28 ele/m3

drift velocity vd = I/nqA    ............. (1)

     where, cross sectional area A = y1z1 = 1.8346*10-5 m2

substitute the given data in eq (1), we get

       vd = 3.494*10-3 m/s

b)

from Hall effet,

electric field E = vdB    ........... (2)

substitute the given data in eq (2), we get

         E = 7.6868*10-4 V/m

c)

from right hand rule, we have magnetic force on

electrons are in positive Z-direction.

therefore, the electric field will be positive Z-direction.

d)

Hall emf = Ez1

                = (7.6868*10-4)(13.49*10-3 )

               = 10.37*10-6 V

                = 10.37 ?V

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