A straight conductor of uniform cross-section carries a current i. If s is the specific charge of an electron, the momentum of all the free electrons per unit length of the conductor, due to their drift velocity only is

Question

A straight conductor of uniform cross-section carries a current i. If s is the specific charge of an electron, the momentum of all the free electrons per unit length of the conductor, due to their drift velocity only is (A) is (B) √i / s (C) i / s (D) (i / s)2

Tardigrade Admin · Accepted Answer

i=n Aev d Or vd=(i/n A e ) Total number of free electrons in the unit length of conductor, N=n A × 1 Total linear momentum of all the free electrons per unit length =( Nm ) v d= nAm × (i/ nAe )=(i/(e/m)) =(i/s)

Q. A straight conductor of uniform cross-section carries a current $i$ . If $s$ is the specific charge of an electron, the momentum of all the free electrons per unit length of the conductor, due to their drift velocity only is

is

$i / s$

$i / s$

$(i / s)^{2}$

$i = n A e v_{d}$
Or $v_{d} = \frac{i}{n A e}$
Total number of free electrons in the unit length of conductor,
$N = n A \times 1$
Total linear momentum of all the free electrons per unit length
$= (N m) v_{d} = n A m \times \frac{i}{n A e} = \frac{i}{( e / m )} = \frac{i}{s}$

Q. A straight conductor of uniform cross-section carries a current i. If s is the specific charge of an electron, the momentum of all the free electrons per unit length of the conductor, due to their drift velocity only is

is

i/s​

i/s

(i/s)2

i=nAevd​ Or vd​=nAei​ Total number of free electrons in the unit length of conductor, N=nA×1 Total linear momentum of all the free electrons per unit length =(Nm)vd​=nAm×nAei​=(e/m)i​=si​

Q. A straight conductor of uniform cross-section carries a current $i$ . If $s$ is the specific charge of an electron, the momentum of all the free electrons per unit length of the conductor, due to their drift velocity only is

$i / s$

$i / s$

$(i / s)^{2}$

$i = n A e v_{d}$
Or $v_{d} = \frac{i}{n A e}$
Total number of free electrons in the unit length of conductor,
$N = n A \times 1$
Total linear momentum of all the free electrons per unit length
$= (N m) v_{d} = n A m \times \frac{i}{n A e} = \frac{i}{( e / m )} = \frac{i}{s}$