Physics Questions and Answers – Current Electricity – Drift of Electrons and the Origin of Resistivity

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This set of Physics Assessment Questions for Schools focuses on “Current Electricity – Drift of Electrons and the Origin of Resistivity”.

1. What is the SI unit of mobility?
a) Vm-1
b) m2V-1s-1
c) mV-2
d) m2V-2s-1
View Answer

Answer: b
Explanation: The SI unit of mobility is m2V-1s-1.
Mobility = \(\frac {Drift \, velocity}{Electric \, field}\)
SI unit = \(\frac {ms^{-1}}{Vm^{-1}}\) = m2V-1s-1.
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2. Consider a conductor of length 0.5 m. A potential difference of 20V is applied across this conductor. If the drift velocity of electrons is given as 5.0 × 10-4ms-1, then determine the mobility of the electrons.
a) 5.25 × 1018m2V-1s-1
b) 5.25 × 10-18m2V-1s-1
c) 1.25 × 10-5 m2V-1s-1
d) 1.25 × 106m2V-1s-1
View Answer

Answer: c
Explanation: Mobility = \(\frac {Drift \, velocity}{Electric \, field}\).
Electric field = \(\frac {Voltage}{length} = \frac {20}{0.5}\) = 40Vm-1.
Mobility = \(\frac {Drift \, velocity}{Electric \, field} = \frac {5.0 \times 10^{-4}}{40}\) = 1.25 × 10-5 m2V-1s-1.

3. Calculate the drift velocity of free electrons if a current of 5 A is maintained in a conductor of cross-section 10-2m2. The number density of free electrons is 5 × 1020m-3.
a) 6.25 ms-1
b) 5.25 ms-1
c) 2.25 ms-1
d) 12.25 ms-1
View Answer

Answer: a
Explanation: Drift velocity = \(\frac {Current}{(number \, density \, of \, electrons \, \times \, charge \, on \, electron \, \times \, Area)}\).
\(\frac {5}{(5 \times 10^{20} \times 1.6 \times 10^{-19} \times 10^{-2})}\)
= 6.25 ms-1.
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4. Predict the effect of temperature of the conductor on the drift velocity of electrons.
a) Drift velocity varies linearly with temperature
b) Drift velocity does not depend on the temperature
c) Drift velocity increases with increasing temperature
d) Drift velocity decreases with increasing temperature
View Answer

Answer: d
Explanation: On increasing the temperature of a conductor, the value of resistivity of its material increases. Resistivity is indirectly proportional to drift velocity. Therefore, the drift velocity of electrons decreases with the increasing temperature of the conductor.

5. A potential difference of 100 V is applied across a conductor of length 50cm. Calculate the drift velocity of electrons if the electron mobility is 9 × 10-5 m2V-1s-1.
a) 0.001 ms-1
b) 1.800 ms-1
c) 0.018 ms-1
d) 0.180 ms-1
View Answer

Answer: c
Explanation: Drift velocity = Mobility × Electric field.
It can be rearranged as,
Drift velocity = \(\frac {(mobility \, \times \, Potential \, difference)}{length}\).
Drift velocity = \(\frac {(9 × 10^{-5} \times 100)}{0.5}\) = 0.018 ms-1.
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6. The mobility of a charge carrier is the drift velocity acquired by it in a unit cross-sectional area.
a) True
b) False
View Answer

Answer: b
Explanation: The mobility of a charge carrier is the drift velocity acquired by it in a unit electric field.
Mobility=\(\frac {Drift \, velocity}{Electric \, field}\). SI unit = \(\frac {ms^{-1}}{Vm^{-1}}\) = m2V-1s-1.

7. Predict the effect of length of conductor on drift velocity of electrons.
a) Drift velocity varies linearly with the length of conductor
b) Drift velocity does not depend on the length of conductor
c) Drift velocity increases with the increasing length of conductor
d) Drift velocity decreases with the increasing length of conductor
View Answer

Answer: d
Explanation: The drift velocity of electrons decreases when the length of the conductor is increased.
Drift velocity = \(\frac {Potential \, difference}{(number \, of \, electrons \, \times \, charge \, of \, electron \, \times \, length \, of \, conductor \, \times \, density)}\).
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8. Calculate the drift velocity of free electrons if a current of 200 A is maintained in a conductor of cross-section 10-3m2. The number density of free electrons is 96 × 1020 m-3.
a) 130.20 ms-1
b) 13.020 ms-1
c) 1.3020 ms-1
d) 0.1302 ms-1
View Answer

Answer: a
Explanation: Drift velocity = \(\frac {Current}{(number \, density \, of \, electrons \, \times \, charge \, on \, electron \, \times \, Area)}\).
=\(\frac {200}{(96 \times 10^{20} \times 1.6 \times 10^{-19} \times 10^{-3})}\)
= 130.20 ms-1.

Sanfoundry Global Education & Learning Series – Physics – Class 12.

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