Electrons are emitted by a hot filament and are accelerated by an electric field, as shown in the figure. The two stops at the left ensure that the electron beam has a uniform cross-section.enter image description here

(a) The speed of the electrons is more at B than at A.

(b) The electric current is from left to right.

(c) The magnitude of the current is larger at B than at A.

(d) The current density is more at B than at A.

The answer is only a). However, I think c) and d) are also correct.

Thus, my question is if the speed of electron (in space) is increased, will the magnitude of current also increase?

Question taken from Concept of physics by HC Verma part 2 chapter current electricity objective 2 question 1.


closed as off-topic by John Rennie, Thomas Fritsch, WillO, tpg2114 Jun 30 at 11:19

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They clearly say to consider the electron beam as having uniform cross section. The current cannot increase from A to B because this will imply some charge is created between the two points. You can think the two points as belonging to a series circuit. Then if the current does not change and the cross section area does not change the current density should be the same. This is all you need to answer the quiz.

But then, you may wonder how to reconcile this with the formula for current density, $j=nev$. If the velocity increases, how come that j does not? The answer is that n, the electron density decreases as the electrons speed up. Think about two electrons emitted one second apart. If they move with 1 cm/s, there will be 1cm distance between them. If they move 2cm/s, they will be 2 cm apart, so the density decreases.


OK, I think that c and d cannot be decided as correct, because together with the higher speed the area with electrons becomes larger , the opening at A makes it a point source.

Definition of current :


Definition of current density:

$j=$ limit as $A$ goes to zero of $I/A$

So I think more geometric information is needed in order to decide whether c and d are also correct, but definitely A is correct.

  • $\begingroup$ Why use the concept of drift velocity in free space? $\endgroup$ – user541396 Jun 29 at 13:09
  • $\begingroup$ @user541396 I assume that in free space v describes the velocity of the electrons, i.e. they" drift" towards the field and generate the I . as they do not find atoms to stop them the velocity increases $\endgroup$ – anna v Jun 29 at 13:15

No, actually it will decrease as, from $$ Q=ne \tag{1} $$ and $$ I=q/t. \tag{2} $$ From (1) and (2) we have, $$ I=n \cdot e \cdot t $$ So, as the speed increases the time taken by electron decreases, so the current will also decrease (taking $n$ and $e$ as constants)

  • $\begingroup$ But $ v_d $ will increase. $\endgroup$ – user541396 Jun 29 at 17:08
  • $\begingroup$ I have improved your formatting by using MathJax. However, I can't follow your conclusion to reaach the third equation. Shouldn't it be $I = n \cdot e / t$ ? $\endgroup$ – Thomas Fritsch Jun 29 at 17:23

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