
Drift velocity varies with intensity of electric field as per the relation
(A) ${v_d} \propto E$
(B) ${v_d} \propto \dfrac{1}{E}$
(C) ${v_d} = cons\tan t$
(D) ${v_d} \propto {E^2}$
Answer
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Hint:The average velocity with which the free electrons move in a current carrying metal wire is called drift velocity of electrons (vd).Subatomic particles like electrons move in random directions all the time. When electrons are subjected to an electric field they do move randomly, but they slowly drift in one direction, in the direction of the electric field applied. The net velocity at which these electrons drift is known as drift velocity.
Formula used:
So, ${v_d} = \dfrac{I}{{neA}}$
Where, n = number density of electrons
e = charge of an electron
I = electric current through the wire
A = cross-sectional area of the wire
Complete step by step answer:
From the definition of drift velocity we know, ${v_d} = \dfrac{I}{{neA}}$
Suppose, potential difference of V is applied at two ends of homogeneous conductor of length l.So, the magnitude of uniform electric field produced inside the conductor is,
$E = \dfrac{V}{l}$
So, the force acting on a free electron inside the conductor, $eE = \dfrac{{eV}}{l}$
Due to this acceleration the velocity of the electron will continuously increase, but practically it does not happen. The motion of electrons is thwarted by the collision of atoms and ions inside the conductor. If the opposing force I considered to be equivalent to a viscous force then the force acting against the motion of the electron inside the metallic conductor is proportional to velocity of electron.
Opposing force = $k{v_d}$where, k is a constant
So, condition of equilibrium is
$
eE = k{v_d} \\
or,{v_d} = \dfrac{{eE}}{k} = \dfrac{e}{k}E \\
$
From the above relation it is clear that, ${v_d} \propto E$ as $\dfrac{e}{k}$ is a constant
So, the correct answer is option (A) ${v_d} \propto E$.
Note:The constant $\mu = \dfrac{e}{k} = \dfrac{{{v_d}}}{E}$ is called mobility of free electrons.Mobility of a free electron is the uniform drift velocity attained by it due to application of unit uniform electric field inside metallic conductor.
Formula used:
So, ${v_d} = \dfrac{I}{{neA}}$
Where, n = number density of electrons
e = charge of an electron
I = electric current through the wire
A = cross-sectional area of the wire
Complete step by step answer:
From the definition of drift velocity we know, ${v_d} = \dfrac{I}{{neA}}$
Suppose, potential difference of V is applied at two ends of homogeneous conductor of length l.So, the magnitude of uniform electric field produced inside the conductor is,
$E = \dfrac{V}{l}$
So, the force acting on a free electron inside the conductor, $eE = \dfrac{{eV}}{l}$
Due to this acceleration the velocity of the electron will continuously increase, but practically it does not happen. The motion of electrons is thwarted by the collision of atoms and ions inside the conductor. If the opposing force I considered to be equivalent to a viscous force then the force acting against the motion of the electron inside the metallic conductor is proportional to velocity of electron.
Opposing force = $k{v_d}$where, k is a constant
So, condition of equilibrium is
$
eE = k{v_d} \\
or,{v_d} = \dfrac{{eE}}{k} = \dfrac{e}{k}E \\
$
From the above relation it is clear that, ${v_d} \propto E$ as $\dfrac{e}{k}$ is a constant
So, the correct answer is option (A) ${v_d} \propto E$.
Note:The constant $\mu = \dfrac{e}{k} = \dfrac{{{v_d}}}{E}$ is called mobility of free electrons.Mobility of a free electron is the uniform drift velocity attained by it due to application of unit uniform electric field inside metallic conductor.
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