Question

If resishvity of copper conductor is 1.7 x 10^8 and electric field
is loo v/m then current density
will be​

Answer 1

Answer:

Solve Further to get The desired Answer.

Thanks.

Answer 2

Answer: 5.8 x 10⁻⁷ A/m²

We use the microscopic form of Ohm's Law

which is J = σE                      { Where J = Current Density

                                                             σ = Conductivity

                                                             E = Electric field }

To derive this,

We know, the Drift velocity of an electron:  Vd = $\frac{e E T}{m}$  ---(1)  { e = Charge of electron

                                                                                             E = Electric field

                                                                                             T = Relaxation Time          

                                                                                             m = Mass of electron}

and the Drift Current is I = n e A Vd  -----(2)     {A = Area of cross section}

$\frac{I}{A}$ = n e Vd

substituting the value of Vd from (1),

$\frac{I}{A}$ = n e $\frac{e E T}{m}$

And we know, $\frac{I}{A}$  = J,

∴ J = n e² E $\frac{T}{m}$  ------------(3)

we know that ρ = $\frac{m}{n e^{2}T }$        { Can be derived in the same way}

and σ =  1/ρ

Hence, σ = $\frac{ne^{2}T }{m}$

∴ Eqn (3) becomes

J = σE

or J = 1/ρ x E

Substituting the values,

∴ J =  $\frac{1}{1.7 * 10^{8} }$ x 100

∴ J =  5.8 x 10⁻⁷ A/m²