Question

the drift velocity of free electrons in a conductor is v,the current i is flowing in it.if both the radius and current are doubled then drift velocity will be?

Answer 1

hello friend...!!


• According to the question, the radius and the current is doubled

we know, drift velocity is equals to,

Vd = I / e. n. a

a = πR^2

implies,

Vd = 2I / π(2R) ^2 .e. n


Vd = 2I / 4πR^2 .e. n

implies,

Vd = I / 2πR^2 .e. n

therefore,

Vd = I / 2e. n. a

therefore the drift velocity is reduced to half.

________________________________

Hope it helps..!!

Answer 2

Answer:

If both the radius and current are doubled then drift velocity will become half of the initial value.

Explanation:

The drift velocity is calculated as,

$v_d=\frac{I}{ne\times \pi r^2}$      (1)

Where,

$v_d$=drift velocity

I=current flowing through a cross-section

n=concentration of an electron

e=charge on an electron

πr² =area of the cross-section     (r=radius)

Since both radius and current are doubled then equation (1) becomes,

$v_d'=\frac{2I}{ne\times \pi (2r)^2}$

$v_d'=\frac{2I}{ne\times \pi \times 4r^2}$

$v_d'=\frac{I}{ne\times \pi \times 2r^2}$      (2)

On comparing equations (1) and (2) we get;

$v_d'=\frac{v_d}{2}$      (3)

Hence, if both the radius and current are doubled then drift velocity will become half of the initial value.

#SPJ2