A potential difference is applied across the ends of a metallic wire. If the potential difference is doubled, the drift velocity will be
A. Be Doubled
B. Be halved
C. Be quadrupled
D. Remain unchanged
Answer
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Hint: Drift velocity is defined as the velocity with which the free electrons get drifted towards the positive terminal from negative terminal under the effect of applied external field. It is the average speed of movement of electrons inside conductors.
Formula to be used: The drift velocity of electrons inside a conductor is given by
${{v}_{d}}=\dfrac{eV}{mL}\tau $
Where,
${{v}_{d}}$ is the drift velocity.
$e$ is the charge on an electron.
$V$ is the applied potential difference along the conductor.
$m$ is the mass of an electron.
$L$ is the length of the conductor.
$\tau $ is the average time of motion of an electron inside the conductor.
Complete step-by-step solution:
When two ends of a conductor are joined to a battery then one end is at higher potential and the other is at lower potential. This produces an electric field inside the conductor from a point of higher potential to lower potential which is given by, $E=\dfrac{V}{L}$. The field exerts an electric force $\left( F=eE \right)$ on free electrons causing acceleration of each electron, $a=\dfrac{F}{m}=\dfrac{eE}{m}$. So, the drift velocity of electron is given by,
$\begin{align}
& {{v}_{d}}=a\times \tau =\dfrac{eE}{m}\tau =\dfrac{eV}{mL}\tau \\
& \Rightarrow {{v}_{d}}=KV\text{ }\left( K=\dfrac{e}{mL}\tau \text{ is constant} \right) \\
\end{align}$
We can see that drift velocity ${{v}_{d}}$ is directly proportional to applied potential difference across the conductor. So, if the applied potential difference is doubled then the drift velocity of electrons will also get doubled.
Hence, the correct answer is $\left( \text{A} \right)$.
Note: The current flowing in a conductor is directly proportional to the drift velocity of electrons. The electrons inside a conductor move with random velocities in random direction until an electric field is applied to the conductor. The electric field accelerates the electrons towards the positive terminal of the battery.
Formula to be used: The drift velocity of electrons inside a conductor is given by
${{v}_{d}}=\dfrac{eV}{mL}\tau $
Where,
${{v}_{d}}$ is the drift velocity.
$e$ is the charge on an electron.
$V$ is the applied potential difference along the conductor.
$m$ is the mass of an electron.
$L$ is the length of the conductor.
$\tau $ is the average time of motion of an electron inside the conductor.
Complete step-by-step solution:
When two ends of a conductor are joined to a battery then one end is at higher potential and the other is at lower potential. This produces an electric field inside the conductor from a point of higher potential to lower potential which is given by, $E=\dfrac{V}{L}$. The field exerts an electric force $\left( F=eE \right)$ on free electrons causing acceleration of each electron, $a=\dfrac{F}{m}=\dfrac{eE}{m}$. So, the drift velocity of electron is given by,
$\begin{align}
& {{v}_{d}}=a\times \tau =\dfrac{eE}{m}\tau =\dfrac{eV}{mL}\tau \\
& \Rightarrow {{v}_{d}}=KV\text{ }\left( K=\dfrac{e}{mL}\tau \text{ is constant} \right) \\
\end{align}$
We can see that drift velocity ${{v}_{d}}$ is directly proportional to applied potential difference across the conductor. So, if the applied potential difference is doubled then the drift velocity of electrons will also get doubled.
Hence, the correct answer is $\left( \text{A} \right)$.
Note: The current flowing in a conductor is directly proportional to the drift velocity of electrons. The electrons inside a conductor move with random velocities in random direction until an electric field is applied to the conductor. The electric field accelerates the electrons towards the positive terminal of the battery.
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