Question

The mean drift speed of free electrons in a conductor of radius r is v. If same current flows in a conductor of same material but of radius 2r, the mean drift speed will now be

Answer 1

The mean drift speed will be v/4

Explanation:

We know that

$I=neAv_d$

or, $v_d=\frac{I}{neA}$       . . . . . . . . . . (1)

Where

A is the area of cross section

n is electon density

I is current

$v_d$ is drift speed

If radius of the wire is r then

$A=\pi r^2$

Therefore from (1)

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

If current is constant

$v_d\propto \frac{1}{r^2}$

or. $\frac{v_{d1}}{v_{d2}}=\frac{r_2^2}{r_1^2}$

$\implies \frac{v}{v_{d2}}=\frac{(2r)^2}{r^2}$

$\implies v_{d2}=\frac{v}{4}$

Hope this answer is helpful.

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Answer 2

Answer:

There can be no field in a real-world conductor unless there is a current. There can be no e-field inside a metal object at equilibrium. For electrostatic situations, the movable charges in a conductor will flow briefly, in order to arrange themselves to remove any internal e-fields. Then the conductor has attained electrical equilibrium, and the internal field is then zero. But in electrodynamic situations, of course there is an e-field inside a conductor. If there weren’t, then electric currents couldn’t exist.

Also: there cannot be an e-field inside a perfect conductor unless current is changing. For magneto-static situations (constant current,) the e-field inside a perfect conductor must be zero. But for changing currents and changing magnetic fields, inductance plays a role, and there will be an induced e-field which accelerates charges to create changing currents. Or in other words, for currents to become established inside superconductors, an e-field must briefly exist inside the superconductor.

Inside resistors (such as metal wires,) Ohm’s law at the micro-scale says that current density is proportional to the e-field and inversely proportional to the resistivity of the material. In other words, the higher the value of e-field inside a conductor, the faster is the conductor’s charge-carriers will drift along. Double the voltage along a wire and we double the drift speed, which doubles the amperes. And for zero e-fields, on average the carriers stop drifting, so the value of amperes is zero.