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A conductor with cross-sectional area of 10 cm2 carrier a conductor current 2sin (109 t) mA. If σ = 2.5 x 106 S/m and εr 4.6 the magnitude of the displacement current density is
- a)48.4 μA/m2
- b)8.11 nA/m2
- c)32.6 nA/m2
- d)16.4 μA/m2
Correct answer is option 'C'. Can you explain this answer?
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Given:
Cross-sectional area of the conductor (A) = 10 cm^2 = 10 × 10^(-4) m^2
Conductor current (I) = 2sin(109t) mA = 2 × 10^(-3)sin(109t) A
Conductivity (σ) = 2.5 × 10^6 S/m
Distance from the conductor (r) = 4.6 m
To find: Magnitude of the displacement current density (Jd)
Formula:
Displacement current density (Jd) = ε0(dE/dt)
Where:
ε0 = Permittivity of free space = 8.854 × 10^(-12) F/m
dE/dt = Rate of change of electric field with respect to time
Calculation:
1. Calculating the rate of change of electric field (dE/dt):
Given that the conductor current is given by I = 2sin(109t) A.
So, the magnetic field (B) around the conductor can be calculated using Ampere's law:
∮ B·dl = μ0I
B × 2πr = μ0(2sin(109t))
B = (μ0/2πr)(2sin(109t))
Differentiating both sides with respect to time (t):
d(B)/dt = (μ0/2πr)(2cos(109t))(109)
d(B)/dt = (μ0/πr)(109cos(109t))
Now, using Faraday's law of electromagnetic induction:
E = -d(B)/dt
E = -(μ0/πr)(109cos(109t))
Therefore, dE/dt = (d/dt)(-(μ0/πr)(109cos(109t)))
dE/dt = (109μ0/πr)(sin(109t))
2. Calculating the displacement current density (Jd):
Using the formula Jd = ε0(dE/dt), we substitute the value of dE/dt calculated above:
Jd = ε0(109μ0/πr)(sin(109t))
Jd = (109ε0μ0/πr)(sin(109t))
Jd = (109 × 8.854 × 10^(-12) × 4π × 10^(-7)/π × 4.6)(sin(109t))
Jd = (109 × 8.854 × 10^(-12) × 4π × 10^(-7)/4.6)(sin(109t))
Jd ≈ 32.6 × 10^(-9) sin(109t) A/m^2
Therefore, the magnitude of the displacement current density is approximately 32.6 nA/m^2. Hence, option C is correct.
Cross-sectional area of the conductor (A) = 10 cm^2 = 10 × 10^(-4) m^2
Conductor current (I) = 2sin(109t) mA = 2 × 10^(-3)sin(109t) A
Conductivity (σ) = 2.5 × 10^6 S/m
Distance from the conductor (r) = 4.6 m
To find: Magnitude of the displacement current density (Jd)
Formula:
Displacement current density (Jd) = ε0(dE/dt)
Where:
ε0 = Permittivity of free space = 8.854 × 10^(-12) F/m
dE/dt = Rate of change of electric field with respect to time
Calculation:
1. Calculating the rate of change of electric field (dE/dt):
Given that the conductor current is given by I = 2sin(109t) A.
So, the magnetic field (B) around the conductor can be calculated using Ampere's law:
∮ B·dl = μ0I
B × 2πr = μ0(2sin(109t))
B = (μ0/2πr)(2sin(109t))
Differentiating both sides with respect to time (t):
d(B)/dt = (μ0/2πr)(2cos(109t))(109)
d(B)/dt = (μ0/πr)(109cos(109t))
Now, using Faraday's law of electromagnetic induction:
E = -d(B)/dt
E = -(μ0/πr)(109cos(109t))
Therefore, dE/dt = (d/dt)(-(μ0/πr)(109cos(109t)))
dE/dt = (109μ0/πr)(sin(109t))
2. Calculating the displacement current density (Jd):
Using the formula Jd = ε0(dE/dt), we substitute the value of dE/dt calculated above:
Jd = ε0(109μ0/πr)(sin(109t))
Jd = (109ε0μ0/πr)(sin(109t))
Jd = (109 × 8.854 × 10^(-12) × 4π × 10^(-7)/π × 4.6)(sin(109t))
Jd = (109 × 8.854 × 10^(-12) × 4π × 10^(-7)/4.6)(sin(109t))
Jd ≈ 32.6 × 10^(-9) sin(109t) A/m^2
Therefore, the magnitude of the displacement current density is approximately 32.6 nA/m^2. Hence, option C is correct.
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A conductor with cross-sectional area of 10 cm2carrier a conductor current 2sin (109 t) mA. Ifσ =2.5 x 106 S/m andεr 4.6the magnitude of the displacement current density isa)48.4μA/m2b)8.11 nA/m2c)32.6 nA/m2d)16.4 μA/m2Correct answer is option 'C'. Can you explain this answer?
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A conductor with cross-sectional area of 10 cm2carrier a conductor current 2sin (109 t) mA. Ifσ =2.5 x 106 S/m andεr 4.6the magnitude of the displacement current density isa)48.4μA/m2b)8.11 nA/m2c)32.6 nA/m2d)16.4 μA/m2Correct answer is option 'C'. Can you explain this answer? for Electronics and Communication Engineering (ECE) 2024 is part of Electronics and Communication Engineering (ECE) preparation. The Question and answers have been prepared according to the Electronics and Communication Engineering (ECE) exam syllabus. Information about A conductor with cross-sectional area of 10 cm2carrier a conductor current 2sin (109 t) mA. Ifσ =2.5 x 106 S/m andεr 4.6the magnitude of the displacement current density isa)48.4μA/m2b)8.11 nA/m2c)32.6 nA/m2d)16.4 μA/m2Correct answer is option 'C'. Can you explain this answer? covers all topics & solutions for Electronics and Communication Engineering (ECE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A conductor with cross-sectional area of 10 cm2carrier a conductor current 2sin (109 t) mA. Ifσ =2.5 x 106 S/m andεr 4.6the magnitude of the displacement current density isa)48.4μA/m2b)8.11 nA/m2c)32.6 nA/m2d)16.4 μA/m2Correct answer is option 'C'. Can you explain this answer?.
A conductor with cross-sectional area of 10 cm2carrier a conductor current 2sin (109 t) mA. Ifσ =2.5 x 106 S/m andεr 4.6the magnitude of the displacement current density isa)48.4μA/m2b)8.11 nA/m2c)32.6 nA/m2d)16.4 μA/m2Correct answer is option 'C'. Can you explain this answer? for Electronics and Communication Engineering (ECE) 2024 is part of Electronics and Communication Engineering (ECE) preparation. The Question and answers have been prepared according to the Electronics and Communication Engineering (ECE) exam syllabus. Information about A conductor with cross-sectional area of 10 cm2carrier a conductor current 2sin (109 t) mA. Ifσ =2.5 x 106 S/m andεr 4.6the magnitude of the displacement current density isa)48.4μA/m2b)8.11 nA/m2c)32.6 nA/m2d)16.4 μA/m2Correct answer is option 'C'. Can you explain this answer? covers all topics & solutions for Electronics and Communication Engineering (ECE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A conductor with cross-sectional area of 10 cm2carrier a conductor current 2sin (109 t) mA. Ifσ =2.5 x 106 S/m andεr 4.6the magnitude of the displacement current density isa)48.4μA/m2b)8.11 nA/m2c)32.6 nA/m2d)16.4 μA/m2Correct answer is option 'C'. Can you explain this answer?.
Solutions for A conductor with cross-sectional area of 10 cm2carrier a conductor current 2sin (109 t) mA. Ifσ =2.5 x 106 S/m andεr 4.6the magnitude of the displacement current density isa)48.4μA/m2b)8.11 nA/m2c)32.6 nA/m2d)16.4 μA/m2Correct answer is option 'C'. Can you explain this answer? in English & in Hindi are available as part of our courses for Electronics and Communication Engineering (ECE).
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