The drift velocity of electrons in a metallic conductor having an external voltage across its ends, can be increased by:
(A) increasing the area of cross-section of the conductor.
(B) decreasing the length of the conductor.
(C) increasing the voltage across the ends of the conductor.
(D) increasing the temperature of the conductor.
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → (B) and (C) only
The electric field inside the conductor is E = V/L, and the drift velocity is the average velocity the electrons gain from this field between collisions (\( \tau \) is the relaxation time).
\( \displaystyle v_d = \frac{eE\tau}{m} = \frac{eV\tau}{mL} \)
(A) increasing the area of cross-section of the conductor. — Incorrect. The area does not appear in \( v_d \). A thicker conductor carries a larger current \( I = neAv_d \), but the drift velocity stays the same.
(B) decreasing the length of the conductor. — Correct. For the same voltage a shorter conductor has a stronger field E = V/L, so \( v_d \) increases.
(C) increasing the voltage across the ends of the conductor. — Correct. A larger V gives a stronger field and a larger \( v_d \).
(D) increasing the temperature of the conductor. — Incorrect. At a higher temperature the ions vibrate more, collisions become more frequent and \( \tau \) decreases, so \( v_d \) decreases.