A block in the shape of a rectangular solid has a cross-sectional area of 3.13 c

Question

A block in the shape of a rectangular solid has a cross-sectional area of 3.13 cm2 across its width, a front-to-rear length of 22.4 cm, and a resistance of 965 . The block's material contains 4.20 × 1022 conduction electrons/m3. A potential difference of 33.8 V is maintained between its front and rear faces. (a) What is the current in the block? (b) If the current density is uniform, what is its magnitude? What are (c) the drift velocity of the conduction electrons and (d) the magnitude of the electric field in the block?

-dont round any numbers

Explanation / Answer

Given,

A = 3.13 cm^2 ; L = 22.4 cm ; R = 965 Ohm ; n = 4.2 x 10^22 electrons ; V = 33.8 V

a)From Ohm's law:

V = IR =. I = V/R

I = 33.8/965 = 0.035 A

Hence, I = 0.035 A

b)We know that the current density is given by:

J = I/A

J = 0.035 A /3.13 x 10^-4 m^2 = 111.821 A/m^2

Hence, J = 111.821 A/m^2

c)drift velocity is given by:

Vd = J/n e

Vd = 111.821/(4.2 x 10^22 x 1.6 x 10^-19) = 0.0166 m/s

Hence, Vd = 0.0166 m/s

d)We know that

E = V/d = 33.8/0.224 = 150.893 V/m

Hence, E = 150.893 V/m

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