Answer
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Hint: Use the relation between current and drift velocity. We know that the cross-sectional area of the wire is \[\pi {r^2}\]. The radius is half of the diameter.
Formula used:
\[I = neA{v_d}\]
Here, n is the number of electrons, e is the charge on an electron, A is the area of cross-section of wire and \[{v_d}\] is the drift velocity.
Complete step by step answer:The drift velocity of electrons is the velocity with which the electrons drift back and forth in a wire or material due to an applied electric field.
We know the relation between current and drift velocity,
\[I = neA{v_d}\]
Here, n is the number of electrons, e is the charge on the electron, A is the area of cross-section of wire and \[{v_d}\] is the drift velocity.
The cross-sectional area of wire of radius r is given as,
\[A = \pi {r^2}\]
Therefore, the current flowing through wire will become,
\[I = ne\pi {r^2}{v_d}\]
We have given, the current in the wire is I when the drift velocity is v. therefore,
\[I = ne\pi {r^2}v\]
Or,
\[I = ne\pi {\left( {\dfrac{d}{2}} \right)^2}v\]
\[ \Rightarrow I = \dfrac{{ne\pi {d^2}v}}{4}\]
\[ \Rightarrow v = \dfrac{{4I}}{{ne\pi {d^2}}}\] …… (1)
Here, d is the diameter of the wire.
The same current flows through wire when the diameter is d/2, then the drift velocity of electrons will become,
\[v' = \dfrac{{4I}}{{ne\pi {{\left( {d/2} \right)}^2}}}\]
\[ \Rightarrow v' = 4\left( {\dfrac{{4I}}{{ne\pi {d^2}}}} \right)\]
\[ \Rightarrow v' = 4v\]
Therefore, the drift velocity will become 4 times the initial drift velocity.
So, the correct answer is option (D).
Note:It comes handy to solve such types of questions if you know the proportionality relation between two variables. Since the drift velocity is inversely proportional to square of the diameter of the wire, the drift velocity will surely increase if the diameter decreases. Since the diameter has decreased by the factor d/2, the drift velocity should be increased 4 times since it is inversely proportional to square of the diameter.
Formula used:
\[I = neA{v_d}\]
Here, n is the number of electrons, e is the charge on an electron, A is the area of cross-section of wire and \[{v_d}\] is the drift velocity.
Complete step by step answer:The drift velocity of electrons is the velocity with which the electrons drift back and forth in a wire or material due to an applied electric field.
We know the relation between current and drift velocity,
\[I = neA{v_d}\]
Here, n is the number of electrons, e is the charge on the electron, A is the area of cross-section of wire and \[{v_d}\] is the drift velocity.
The cross-sectional area of wire of radius r is given as,
\[A = \pi {r^2}\]
Therefore, the current flowing through wire will become,
\[I = ne\pi {r^2}{v_d}\]
We have given, the current in the wire is I when the drift velocity is v. therefore,
\[I = ne\pi {r^2}v\]
Or,
\[I = ne\pi {\left( {\dfrac{d}{2}} \right)^2}v\]
\[ \Rightarrow I = \dfrac{{ne\pi {d^2}v}}{4}\]
\[ \Rightarrow v = \dfrac{{4I}}{{ne\pi {d^2}}}\] …… (1)
Here, d is the diameter of the wire.
The same current flows through wire when the diameter is d/2, then the drift velocity of electrons will become,
\[v' = \dfrac{{4I}}{{ne\pi {{\left( {d/2} \right)}^2}}}\]
\[ \Rightarrow v' = 4\left( {\dfrac{{4I}}{{ne\pi {d^2}}}} \right)\]
\[ \Rightarrow v' = 4v\]
Therefore, the drift velocity will become 4 times the initial drift velocity.
So, the correct answer is option (D).
Note:It comes handy to solve such types of questions if you know the proportionality relation between two variables. Since the drift velocity is inversely proportional to square of the diameter of the wire, the drift velocity will surely increase if the diameter decreases. Since the diameter has decreased by the factor d/2, the drift velocity should be increased 4 times since it is inversely proportional to square of the diameter.
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