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Figure E27.49 shows a portion of a silver ribbon with z1 = 11.8 mm and y1 = 0.23 mm, carrying a current of 120 A in the +x-direction. The ribbon lies in a uniform magnetic field, in the y-direction, with magnitude 0.95 T. Apply the simplified model of the Hall effect presented in Section 27.9. If there are 5.85 * 1028 free electrons per cubic meter, find (a) the magnitude of the drift velocity of the electrons in the x-direction; (b) the magnitude and direction of the electric field in the z-direction due to the Hall effect; (c) the Hall emf

Short Answer

Expert verified
  1. The drift velocity of the electrons in the x-direction isvd=4.7×10-3m/s
  2. The magnitude and direction of the electric field in the z-direction due to the Hall effect is Ez=4.5×10-3V/m
  3. The hall emfEHall=5.3×10-5V

Step by step solution

01

Drift velocity

The drift velocity is given by

Vd=Jxn|q|

02

Determine the drift velocity

(a)

The drift velocity is given by

Vd=Jxn|q|

Where,n=5.85×1028m-3

The drift velocity is

Jx=IA=Iy1z1=120A(0.23×10−3m)(0.0118m)=4.42×107A/m2Andvd=4.42×107A/m2(5.85×1028m−3)(1.602×10−19C)=4.7×10−3m/s

Therefore, the drift velocity of the electrons in the x-direction is vd=4.7×10−3m/s

03

Determine the magnitude and direction of electric field

(b)

Magnetic force equals to the electric field

|q|Ez=|q|vdBy

Or,

Ez=VdBy=(4.7×10−3m/s)(0.95T)=4.5×10−3V/m

Therefore, the magnitude and direction of the electric field in the z-direction due to the Hall effect isEz=4.5×10−3V/m

04

Determine the hall emf

(c)

The hall emf is

EHall=Ez1=(4.5×10−3V/m)(0.0118m)=5.3×10−5V

Therefore, the hall emf isEHall=5.3×10−5V

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