A solenoid 2.5 cm in diameter and 30 cm in length has 4800 turns and carries a current of 2.0 A. Part A Calculate the magnetic f
lux through the circular cross-sectional area of the solenoid. Hint: Assume this is a very long solenoid and use the simplified magnetic field formula for an infinite solenoid.
1 answer:
Answer: Ø = 0.034Wb
Explanation:
A solenoid is idealized as a certain number of current loops of diameter d in series.
• The current loops create a magnetic field. This field permeates each loop, ie, there is magnetic flux through each loop.
Determine the number of turns from the length and given winding rate.
• Calculate the strength of the magnetic field produced by the solenoid, then the flux through one turn (winding).
Please find the attached file for the solution.
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Answer:
correct answer is C
Explanation:
The time constant of an RC circuit is
τ = RC
so to find the capacitance
C = τ/ R
C = 2.150 / 5.20 10³
C = 4.13 10⁻⁴ F
to find the error we use the worst case
ΔC = |
the absolute value guarantees that we find the worst case, we evaluate the derivatives
ΔC = 1 /R Δτ + τ/R² ΔR
the absolute values of the errors are
Δτ = 0.002 s
ΔR = 0.3 kΩ
we substitute
ΔC = 0.002 /5.20 10³ + 2.150/(5.20 10³)² 0.3 10³
ΔC = 3.8 10⁻⁷ + 1.74 10⁻⁵
ΔC = 1.77 10⁻⁵ F
the uncertainty or error must be expressed with a significant figure
ΔC = 2 10⁻⁵ F
the percentage error is
Er% =
Er% =
Er% = 4.8%
the correct answer is C