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A wire of diameter 1 mm and conductivity 5 x 107S/m has 1029 free electrons /m3 when an electric field of 10 mV/m is applied.
The current density and the drift velocity of the electrons will be respectively given by (take charge on an electrons e = -1.6 x 10-19 C)
The current density and the drift velocity of the electrons will be respectively given by (take charge on an electrons e = -1.6 x 10-19 C)
- a)500 kA/m2 and 3.125 x 10-5 m/s
- b)300 kA/m2 and 2.225 x 10-5 m/s
- c)500 kA/m2 and 2.225 x 10-5 m/s
- d)300 kA/m2 and 3.125 x 10-5 m/s
Correct answer is option 'A'. Can you explain this answer?
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Verified Answer
A wire of diameter 1 mm and conductivity 5 x 107S/m has 1029 free elec...
The charge density of free electron is:
ρv = ne = (1029) x (-1.6 x 10-19)
= -1.6 x 1010 C/m2
Given,
E = 10 mV/m,
d = 1 mm,
σ = 5 x 107 S/m,
n = 1029/m3
Now, current density,
Also, drift velocity,
ρv = ne = (1029) x (-1.6 x 10-19)
= -1.6 x 1010 C/m2
Given,
E = 10 mV/m,
d = 1 mm,
σ = 5 x 107 S/m,
n = 1029/m3
Now, current density,
Also, drift velocity,
Most Upvoted Answer
A wire of diameter 1 mm and conductivity 5 x 107S/m has 1029 free elec...
Given parameters:
- Diameter of wire = 1 mm
- Conductivity of wire = 5 x 107 S/m
- Free electrons/m3 = 1029
- Electric field = 10 mV/m
- Charge on an electron = -1.6 x 10-19 C
To find:
- Current density (J)
- Drift velocity of electrons (vd)
Formulae used:
- Current density (J) = σE
- Drift velocity of electrons (vd) = μE
- σ = nqμ, where n is the number of free electrons per unit volume, q is the charge on an electron, and μ is the mobility of electrons
Calculation:
- Radius of wire (r) = 0.5 mm = 0.0005 m
- Cross-sectional area of wire (A) = πr2 = π(0.0005)2 = 7.85 x 10-7 m2
- Electric field (E) = 10 mV/m = 0.01 V/m
- Number of free electrons per unit volume (n) = 1029/m3
- Charge on an electron (q) = -1.6 x 10-19 C
- Mobility of electrons (μ) = σ/ (nq) = 5 x 107 / (1029 x -1.6 x 10-19) = -1.953 x 10-5 m2/Vs (Note: The negative sign indicates that the electrons move in the opposite direction of the electric field)
- Current density (J) = σE = 5 x 107 x 0.01 = 500 kA/m2
- Drift velocity of electrons (vd) = μE = -1.953 x 10-5 x 0.01 = -1.953 x 10-7 m/s (Note: The negative sign indicates that the electrons move in the opposite direction of the electric field)
Answer:
The current density and the drift velocity of the electrons will be respectively given by 500 kA/m2 and 3.125 x 10-5 m/s, which is option (A).
- Diameter of wire = 1 mm
- Conductivity of wire = 5 x 107 S/m
- Free electrons/m3 = 1029
- Electric field = 10 mV/m
- Charge on an electron = -1.6 x 10-19 C
To find:
- Current density (J)
- Drift velocity of electrons (vd)
Formulae used:
- Current density (J) = σE
- Drift velocity of electrons (vd) = μE
- σ = nqμ, where n is the number of free electrons per unit volume, q is the charge on an electron, and μ is the mobility of electrons
Calculation:
- Radius of wire (r) = 0.5 mm = 0.0005 m
- Cross-sectional area of wire (A) = πr2 = π(0.0005)2 = 7.85 x 10-7 m2
- Electric field (E) = 10 mV/m = 0.01 V/m
- Number of free electrons per unit volume (n) = 1029/m3
- Charge on an electron (q) = -1.6 x 10-19 C
- Mobility of electrons (μ) = σ/ (nq) = 5 x 107 / (1029 x -1.6 x 10-19) = -1.953 x 10-5 m2/Vs (Note: The negative sign indicates that the electrons move in the opposite direction of the electric field)
- Current density (J) = σE = 5 x 107 x 0.01 = 500 kA/m2
- Drift velocity of electrons (vd) = μE = -1.953 x 10-5 x 0.01 = -1.953 x 10-7 m/s (Note: The negative sign indicates that the electrons move in the opposite direction of the electric field)
Answer:
The current density and the drift velocity of the electrons will be respectively given by 500 kA/m2 and 3.125 x 10-5 m/s, which is option (A).
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A wire of diameter 1 mm and conductivity 5 x 107S/m has 1029 free electrons /m3when an electric field of 10 mV/m is applied.The current density and the drift velocity of the electrons will be respectively given by (take charge on an electrons e = -1.6 x 10-19 C)a)500 kA/m2and 3.125 x 10-5 m/sb)300 kA/m2 and 2.225 x 10-5 m/sc)500 kA/m2 and 2.225 x 10-5 m/sd)300 kA/m2 and 3.125 x 10-5 m/sCorrect answer is option 'A'. Can you explain this answer?
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A wire of diameter 1 mm and conductivity 5 x 107S/m has 1029 free electrons /m3when an electric field of 10 mV/m is applied.The current density and the drift velocity of the electrons will be respectively given by (take charge on an electrons e = -1.6 x 10-19 C)a)500 kA/m2and 3.125 x 10-5 m/sb)300 kA/m2 and 2.225 x 10-5 m/sc)500 kA/m2 and 2.225 x 10-5 m/sd)300 kA/m2 and 3.125 x 10-5 m/sCorrect answer is option 'A'. Can you explain this answer? for Electrical Engineering (EE) 2024 is part of Electrical Engineering (EE) preparation. The Question and answers have been prepared according to the Electrical Engineering (EE) exam syllabus. Information about A wire of diameter 1 mm and conductivity 5 x 107S/m has 1029 free electrons /m3when an electric field of 10 mV/m is applied.The current density and the drift velocity of the electrons will be respectively given by (take charge on an electrons e = -1.6 x 10-19 C)a)500 kA/m2and 3.125 x 10-5 m/sb)300 kA/m2 and 2.225 x 10-5 m/sc)500 kA/m2 and 2.225 x 10-5 m/sd)300 kA/m2 and 3.125 x 10-5 m/sCorrect answer is option 'A'. Can you explain this answer? covers all topics & solutions for Electrical Engineering (EE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A wire of diameter 1 mm and conductivity 5 x 107S/m has 1029 free electrons /m3when an electric field of 10 mV/m is applied.The current density and the drift velocity of the electrons will be respectively given by (take charge on an electrons e = -1.6 x 10-19 C)a)500 kA/m2and 3.125 x 10-5 m/sb)300 kA/m2 and 2.225 x 10-5 m/sc)500 kA/m2 and 2.225 x 10-5 m/sd)300 kA/m2 and 3.125 x 10-5 m/sCorrect answer is option 'A'. Can you explain this answer?.
A wire of diameter 1 mm and conductivity 5 x 107S/m has 1029 free electrons /m3when an electric field of 10 mV/m is applied.The current density and the drift velocity of the electrons will be respectively given by (take charge on an electrons e = -1.6 x 10-19 C)a)500 kA/m2and 3.125 x 10-5 m/sb)300 kA/m2 and 2.225 x 10-5 m/sc)500 kA/m2 and 2.225 x 10-5 m/sd)300 kA/m2 and 3.125 x 10-5 m/sCorrect answer is option 'A'. Can you explain this answer? for Electrical Engineering (EE) 2024 is part of Electrical Engineering (EE) preparation. The Question and answers have been prepared according to the Electrical Engineering (EE) exam syllabus. Information about A wire of diameter 1 mm and conductivity 5 x 107S/m has 1029 free electrons /m3when an electric field of 10 mV/m is applied.The current density and the drift velocity of the electrons will be respectively given by (take charge on an electrons e = -1.6 x 10-19 C)a)500 kA/m2and 3.125 x 10-5 m/sb)300 kA/m2 and 2.225 x 10-5 m/sc)500 kA/m2 and 2.225 x 10-5 m/sd)300 kA/m2 and 3.125 x 10-5 m/sCorrect answer is option 'A'. Can you explain this answer? covers all topics & solutions for Electrical Engineering (EE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A wire of diameter 1 mm and conductivity 5 x 107S/m has 1029 free electrons /m3when an electric field of 10 mV/m is applied.The current density and the drift velocity of the electrons will be respectively given by (take charge on an electrons e = -1.6 x 10-19 C)a)500 kA/m2and 3.125 x 10-5 m/sb)300 kA/m2 and 2.225 x 10-5 m/sc)500 kA/m2 and 2.225 x 10-5 m/sd)300 kA/m2 and 3.125 x 10-5 m/sCorrect answer is option 'A'. Can you explain this answer?.
Solutions for A wire of diameter 1 mm and conductivity 5 x 107S/m has 1029 free electrons /m3when an electric field of 10 mV/m is applied.The current density and the drift velocity of the electrons will be respectively given by (take charge on an electrons e = -1.6 x 10-19 C)a)500 kA/m2and 3.125 x 10-5 m/sb)300 kA/m2 and 2.225 x 10-5 m/sc)500 kA/m2 and 2.225 x 10-5 m/sd)300 kA/m2 and 3.125 x 10-5 m/sCorrect answer is option 'A'. Can you explain this answer? in English & in Hindi are available as part of our courses for Electrical Engineering (EE).
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A wire of diameter 1 mm and conductivity 5 x 107S/m has 1029 free electrons /m3when an electric field of 10 mV/m is applied.The current density and the drift velocity of the electrons will be respectively given by (take charge on an electrons e = -1.6 x 10-19 C)a)500 kA/m2and 3.125 x 10-5 m/sb)300 kA/m2 and 2.225 x 10-5 m/sc)500 kA/m2 and 2.225 x 10-5 m/sd)300 kA/m2 and 3.125 x 10-5 m/sCorrect answer is option 'A'. Can you explain this answer?, a detailed solution for A wire of diameter 1 mm and conductivity 5 x 107S/m has 1029 free electrons /m3when an electric field of 10 mV/m is applied.The current density and the drift velocity of the electrons will be respectively given by (take charge on an electrons e = -1.6 x 10-19 C)a)500 kA/m2and 3.125 x 10-5 m/sb)300 kA/m2 and 2.225 x 10-5 m/sc)500 kA/m2 and 2.225 x 10-5 m/sd)300 kA/m2 and 3.125 x 10-5 m/sCorrect answer is option 'A'. Can you explain this answer? has been provided alongside types of A wire of diameter 1 mm and conductivity 5 x 107S/m has 1029 free electrons /m3when an electric field of 10 mV/m is applied.The current density and the drift velocity of the electrons will be respectively given by (take charge on an electrons e = -1.6 x 10-19 C)a)500 kA/m2and 3.125 x 10-5 m/sb)300 kA/m2 and 2.225 x 10-5 m/sc)500 kA/m2 and 2.225 x 10-5 m/sd)300 kA/m2 and 3.125 x 10-5 m/sCorrect answer is option 'A'. Can you explain this answer? theory, EduRev gives you an
ample number of questions to practice A wire of diameter 1 mm and conductivity 5 x 107S/m has 1029 free electrons /m3when an electric field of 10 mV/m is applied.The current density and the drift velocity of the electrons will be respectively given by (take charge on an electrons e = -1.6 x 10-19 C)a)500 kA/m2and 3.125 x 10-5 m/sb)300 kA/m2 and 2.225 x 10-5 m/sc)500 kA/m2 and 2.225 x 10-5 m/sd)300 kA/m2 and 3.125 x 10-5 m/sCorrect answer is option 'A'. Can you explain this answer? tests, examples and also practice Electrical Engineering (EE) tests.
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