NEET-XII-Physics
32: Electric Current in Conductors
- #1Electrons are emitted by a hot filament and are accelerated by an electric field, as shown in the figure (32-Q1). The two stops at the left ensure that the electron beam has a uniform cross-section.
(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.
Figure
digAnsr: aAns : (a) The speed of the electrons is more at B than at A.
Let the potentials at A and B be VA and VB.
As potential, `` E=-\frac{dV}{dr}`` ,
potential increases in the direction opposite to the direction of the electric field.
Thus, VA < VB
Potential energy of the electrons at points A and B:
UA = `` -``eVA
UB = `` -``eVB
Thus, UA > UB
Let the kinetic energy of an electron at points A and B be KA and KB respectively.
Applying the principle of conservation of mechanical energy, we get:
UA + KA = UB + KB
As, UA > UB,
KA < KB
Therefore, the speed of the electrons is more at B than at A.
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