In a wire of cross-section radius r, free ele | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) The current $I$ in a wire is given by the formula $I = n e A v_d$, where $n$ is the number density of free electrons, $e$ is the charge of an elec

Vedclass · Accepted Answer

$I/2$

Vedclass · Answer

(C) The current $I$ in a wire is given by the formula $I = n e A v_d$, where $n$ is the number density of free electrons, $e$ is the charge of an electron, $A$ is the cross-sectional area, and $v_d$ is the drift velocity.For the first wire: $I_1 = n e (\pi r^2) v$.For the second wire: $r_2 = r/2$ and $v_{d2} = 2v$. The area $A_2 = \pi (r/2)^2 = \pi r^2 / 4 = A_1 / 4$.Thus, $I_2 = n e A_2 v_{d2} = n e (A_1 / 4) (2v) = (1/2) n e A_1 v$.Substituting $I_1 = n e A_1 v$, we get $I_2 = I_1 / 2 = I/2$.

In a wire of cross-section radius $r,$ free electrons travel with drift velocity $v$ when a current $I$ flows through the wire. What is the current in another wire of half the radius and of the same material when the drift velocity is $2v$?

Similar Questions

The general solution of the equation $(\sqrt{3} - 1)\sin \theta + (\sqrt{3} + 1)\cos \theta = 2$ is

$A$ particle is projected from the ground with an initial speed of $v$ at an angle of projection $\theta$. The average velocity of the particle between its time of projection and the time it reaches the highest point of its trajectory is

The tension $T$ in the string shown in the figure is:

The phosphate of a metal has the formula $MHPO_4$. The formula of its chloride would be

$A$ magnetized wire of magnetic moment $M$ and length $l$ is bent into the form of a semicircle of radius $r$. Then its new magnetic moment is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module