Field, force, EMI, transformer MCQs
Practice Field, force, EMI, transformer multiple-choice questions from Magnetic Effects of Electric Current (Class 10 Science) - tap an answer for instant feedback and a step-by-step solution. Practice the full set free on the RankByte app.
Field, force, EMI, transformerQuiz - Solve & Score
Q1. An AC generator's coil rotates at 50 rev/s. Doubling the rotation speed doubles which of the following?
- A.Both the peak EMF and the frequency
- B.Only the frequency
- C.Only the peak EMF
- D.Neither
Answer: A. Both the peak EMF and the frequency
Given 50, asked for the unknown. By EMF_peak = NABω, frequency f = ω/2π. EMF_peak = NABω, frequency f = ω/2π. Both scale linearly with ω, so doubling ω doubles both. Putting it together option A) Both the peak EMF and the frequency.
Q2. Which of the following devices does NOT use the principle of electromagnetic induction?
- A.Electric heater
- B.Transformer
- C.AC generator
- D.Induction coil
Answer: A. Electric heater
That leaves only option A). Confirm with the chapter rule: An electric heater simply uses the heating effect of current (I²R) - no induction involved. Transformer relies on mutual induction; AC generator on rotating coil in a field; induction coil on transient EMF (physics, chapter 'Magnetic Effects of Electric Current'). Answer: A) Electric heater.
Q3. A power-line technician measures the field beside an overhead cable carrying 5 A; what is the magnetic field at a point 0.1 m to the side of the cable?
- A.1*10^-5 T
- B.2*10^-5 T
- C.4*10^-9 T
- D.5*10^-6 T
Answer: A. 1*10^-5 T
For a long straight wire B = mu0*I/(2*pi*r) with mu0 = 4*pi*10^-7. B = (4*pi*10^-7 * 5)/(2*pi*0.1) = (2*10^-7*5)/0.1 = 1*10^-5 T T.
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