Question

A block in the shape of a rectangular solid has a cross-sectional area of 4.84 cm2 across its width, a front-to-rear length of 14.0 cm, and a resistance of 1250 Ω. The block's material contains 6.90 × 1022 conduction electrons/m3. A potential difference of 28.4 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?

Answer 1

(a) I = 0.023 A

(b) J =
`46.9\ A/m^(2)`

(c)
`v_(d) = 2.88* 10^(- 7)\ m/s`

(d) E = 202.85 N/C

Solution:

As per the question:

Cross-sectional area, A =
`4.84\ cm^(2) = 4.84* 10^(- 4)\ m^(2)`

Length of the block, L = 14.0 cm = 0.14 m

R = 1250
`\Omega`

Potential Difference, V =
`28.4\ V`

N =
`6.90* 10^(22)\ conduction\ electrons/m^(3)`

Now,

(a) Current in the block is given by:

`I = (V)/(R)`

`I = (28.4)/(1250) = 0.023\ A`

(b) Current density is given by:

`J = (I)/(A) = (V)/(AR) = (28.4)/(4.84* 10^(- 4)* 1250) = 46.9\ A/m^(2)`

(c) Drift velocity is given by:

`v_(d) = (J)/(ne) = (ALJ)/(Ne) = (VAL)/(NeRA) = (VL)/(NeR)`

where

n =
`(1)/(AL)`

`v_(d) = (28.4* 0.14)/(6.90* 10^(22)* 1.6* 10^(- 19)* 1250) = 2.88* 10^(- 7)\ m/s`

(d) The magnitude of electric field is given by:

`E = (V)/(L) = (28.4)/(0.14) = 202.85\ N/C`

Answer 2

Answer: (a)
`I=0.0227 A`

(b)
`J=46.9422 A.m^(-2)`

(c)
`v_d= 0.0042 m.s^(-1)`

(d)
`E= 202.8571 V.m^(-1)`

Step-by-step explanation:

Given is a solid cube having following properties:

• cross-sectional area, a =
  `4.84 cm^2`
• length of the cube perpendicular to the given area,
  `l= 14 cm`

• resistance,
  `R=1250 \Omega`

• potential difference across the length, V = 28.4 V

• electron density,
  `N= 6.9* 10^(22)  m^(-3)`

We, know the formula:

• Current,
  `I=(V)/(R)`

Putting the respective values:

`I= (28.4)/(1250) \\\\I=0.0227 A`

• Current density,
  `J= (I)/(a)`

Putting the respective values:

`J=(0.0227)/(4.84* 10^(-4))\\\\J=46.9422 A.m^(-2)`

• Drift velocity,
  `v_d= (I)/(N.e.a)`

where, e is the charge on an electron.

Putting the respective values:

`v_d= (0.0227)/(6.9* 10^(22)* 1.6* 10^(-19)* 4.84* 10^(-4))`

`v_d= 0.0042 m.s^(-1)`

Electric field,
`E= (V)/(l)`

`\Rightarrow E=(28.4)/(14* 10^(-2)) \\\\\Rightarrow E= 202.8571 V.m^(-1)`