Flux density calculation in magnetics: Given a magnetic flux of 9 µWb uniformly distributed through a cross-sectional area of 5 × 10^–3 m^2, compute the flux density B and choose the correct value.
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A1800 µT
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B555.56 T
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C9 µT
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D5 × 10^–3 T
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E—
Answer
Correct Answer: 1800 µT
Explanation
Introduction / Context:Flux density B quantifies how much magnetic flux passes through a given area and is crucial for core design, sensor calibration, and avoiding saturation in magnetic components. This problem practices converting micro-units and applying the fundamental relation B = phi / A.
Given Data / Assumptions:
- Total flux, phi = 9 µWb = 9 × 10^–6 Wb.
- Area, A = 5 × 10^–3 m^2.
- Uniform flux distribution across the stated area.
Concept / Approach:The definition of flux density is B = phi / A. After computing in SI units (Wb and m^2), express the result in tesla and convert to more convenient microtesla if appropriate. Careful unit handling prevents order-of-magnitude mistakes.
Step-by-Step Solution:
Write formula: B = phi / A.Insert values: B = (9 × 10^–6) / (5 × 10^–3).Compute: 9/5 = 1.8 and 10^–6 / 10^–3 = 10^–3, so B = 1.8 × 10^–3 T.Convert to microtesla: 1.8 × 10^–3 T = 1.8 mT = 1800 µT.Verification / Alternative check:Check magnitude: Micro-webers over milli-square-meters gives milli-tesla range, which matches 1.8 mT. Unit consistency confirms the calculation.
Why Other Options Are Wrong:
- 555.56 T: Off by orders of magnitude; would require enormous flux or tiny area.
- 9 µT: Would correspond to 9 × 10^–6 Wb spread over 1 m^2, not 5 × 10^–3 m^2.
- 5 × 10^–3 T: Equals 5 mT, not supported by the numbers.
Common Pitfalls:Mishandling micro (10^–6) and milli (10^–3) prefixes; forgetting to divide by area; mixing up webers and tesla definitions.
Final Answer:1800 µT