Faraday’s Law Application: A 100-turn coil has flux per turn φ(t) = (t^3 − 2t) mWb. Find the magnitude of the induced emf at t = 4 s.
Electronics and Communication Engineering
Electromagnetic Field Theory
Difficulty: Easy
Choose an option
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A56 mV
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B46 mV
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C4.6 V
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D5.6 V
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E0.46 V
Answer
Correct Answer: 4.6 V
Explanation
Introduction / Context:Induced electromotive force (emf) in a coil arises from time-varying magnetic flux linkage. This question tests direct application of Faraday’s law to a given flux function, including careful handling of units (mWb to Wb) and differentiation with respect to time.
Given Data / Assumptions:
- Number of turns, N = 100.
- Flux per turn: φ(t) = (t^3 − 2t) mWb.
- Time of interest: t = 4 s.
- Sign convention: magnitude of emf is requested.
Concept / Approach:
Faraday’s law for a coil: e(t) = −N * dφ/dt. Magnitude is |e(t)|. Convert milli-weber to weber when computing volts: 1 mWb = 10^-3 Wb.
Step-by-Step Solution:
Differentiate: dφ/dt = (3t^2 − 2) mWb/s.Evaluate at t = 4 s: dφ/dt = 3*(4^2) − 2 = 48 − 2 = 46 mWb/s.Convert to Wb/s: 46 mWb/s = 46×10^-3 Wb/s.Apply Faraday: |e| = N * |dφ/dt| = 100 * 46×10^-3 = 4.6 V.Verification / Alternative check:
Dimensional check: Wb/s multiplied by turns gives volts; numerical scaling by 10^-3 aligns with milli-prefix, confirming 4.6 V.
Why Other Options Are Wrong:
- 46 mV and 56 mV: off by a factor of 1000 due to ignoring mWb → Wb conversion.
- 5.6 V and 0.46 V: arithmetic/scale errors.
Common Pitfalls:
- Forgetting to convert mWb to Wb.
- Missing the factor of N turns, or confusing flux per turn with total linkage.
Final Answer:
4.6 V