The iqure below shows a portion of a silver ribbon with 2, - 11.8 mm and ys -0.2

Question

The iqure below shows a portion of a silver ribbon with 2, - 11.8 mm and ys -0.23 mm, carrying a current of magnitude 1.00 T. Apply the simplified model of the presented in 101 Ain the .xdrection. The ribbon lies in a uniform magnetic field, in the rdiedon with of the Hall effect presented in Section 27.9. They are 5.85 x 1o free electron per cutic meter a) Find the magritude of the drilt velocity of the electrons in the x direction mýs (b) Find magnitude and the diretion of the electric ield in the 2-drection due to the Hal effect magritude N/C direction direction direction direction (c) Find the Hail emf

Explanation / Answer

Given z1 = 11.8 mm = 0.0118 m

y1 = 0.23 mm = 0.00023 m

Area of cross section A = z1 * y1

A = 0.0118 * 0.00023

A = 2.714 * 10^-6 m^2

current I = 101 A

magnetic field B = 1 T

Charge density n = 5.85 * 10^28 electrons / m^3

a)

Drift velocity v = I / (n * e * A)

v = 101 / (5.85 * 10^28 * 1.6 * 10^-19 * 2.714 * 10^-6)

v = 3.97 * 10^-3 m/s

b)

From the relation

q * E = q * v * B

E = v * B

E = 3.97 * 10^-3 * 1

E = 3.97 * 10^-3 V/m

The direction is -z direction

c)

Hall emf e = E * z1

e = 3.97 * 10^-3 * 0.0118

e = 4.68 * 10^-5 V

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