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

Answer: (a)$I=0.0227 A$

(b) $J=46.9422 A.m^{-2}$

(c) $v_d= 0.0042 m.s^{-1}$

(d) $E= 202.8571 V.m^{-1}$

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, $N= 6.9\times 10^{22}  m^{-3}$

We, know the formula:

• Current, $I=\frac{V}{R}$  

 

Putting the respective values:

$I= \frac{28.4}{1250} \\\\I=0.0227 A$

• Current density, $J= \frac{I}{a}$

Putting the respective values:

$J=\frac{0.0227}{4.84\times 10^{-4}}\\\\J=46.9422 A.m^{-2}$

• Drift velocity, $v_d= \frac{I}{N.e.a}$

where, e is the charge on an electron.

Putting the respective values:

$v_d= \frac{0.0227}{6.9\times 10^{22}\times 1.6\times 10^{-19}\times 4.84\times 10^{-4}}$

$v_d= 0.0042 m.s^{-1}$

Electric field, $E= \frac{V}{l}$

$\Rightarrow E=\frac{28.4}{14\times 10^{-2}} \\\\\Rightarrow E= 202.8571 V.m^{-1}$

Answer 2

Answer:

(a) I = 0.023 A

(b) J = $46.9\ A/m^{2}$

(c) $v_{d} = 2.88\times 10^{- 7}\ m/s$

(d) E = 202.85 N/C

Solution:

As per the question:

Cross-sectional area, A = $4.84\ cm^{2} = 4.84\times 10^{- 4}\ m^{2}$

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

R = 1250$\Omega$

Potential Difference, V = $28.4\ V$

N = $6.90\times 10^{22}\ conduction\ electrons/m^{3}$

Now,

(a) Current in the block is given by:

$I = \frac{V}{R}$

$I = \frac{28.4}{1250} = 0.023\ A$

(b) Current density is given by:

$J = \frac{I}{A} = \frac{V}{AR} = \frac{28.4}{4.84\times 10^{- 4}\times 1250} = 46.9\ A/m^{2}$

(c) Drift velocity is given by:

$v_{d} = \frac{J}{ne} = \frac{ALJ}{Ne} = \frac{VAL}{NeRA} = \frac{VL}{NeR}$

where

n = $\frac{1}{AL}$

$v_{d} = \frac{28.4\times 0.14}{6.90\times 10^{22}\times 1.6\times 10^{- 19}\times 1250} = 2.88\times 10^{- 7}\ m/s$

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

$E = \frac{V}{L} = \frac{28.4}{0.14} = 202.85\ N/C$