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12th Standard Physics — Electromagnetic Induction And Alternating Current: Book Back MCQs with Answers & Explanations

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Every Book Back multiple-choice question from Electromagnetic Induction And Alternating Current (12th Standard Physics, Samacheer Kalvi) — each with the correct option highlighted and a clear, worked explanation. Free to read in English and Tamil.

Answer key at a glance

Q1
An electron moves on a straight line path XY as shown in the figure. The coil abcd is adjacent to the path of the electron. What will be the direction of current, if any, induced in the coil?
An electron moves along a straight line XY near a rectangular conducting loop abcd.
  • A. The current will reverse its direction as the electron goes past the coilCorrect
  • B. No current will be induced
  • C. abcd
  • D. adcb
Explanation. Lenz's law determines the direction of induced current based on the changing magnetic flux created by the moving electron.
Q2
A thin semi-circular conducting ring (PQR) of radius r is falling with its plane vertical in a horizontal magnetic field B, as shown in the figure. The potential difference developed across the ring when its speed v, is
A semicircular conducting loop PQR falls through a uniform horizontal magnetic field B.
  • A. Zero
  • B. \(B \pi rv^2/2\) and P is at higher potential
  • C. \(\pi rBv\) and R is at higher potential
  • D. \(2rBv\) and R is at higher potentialCorrect
Explanation. The motional emf is calculated using the straight-line distance between the endpoints of the conductor moving through the magnetic field.
Q3
The flux linked with a coil at any instant t is given by \(\Phi_B = 10t^2 - 50t + 250\). The induced emf at t = 3 s is
  • A. -190 V
  • B. -10 VCorrect
  • C. 10 V
  • D. 190 V
Explanation. Induced emf is the negative time derivative of the magnetic flux. Differentiating the given equation and substituting time yields the result.
Q4
When the current changes from +2A to -2A in 0.05 s, an emf of 8 V is induced in a coil. The co-efficient of self-induction of the coil is
  • A. 0.2 H
  • B. 0.4 H
  • C. 0.8 H
  • D. 0.1 HCorrect
Explanation. Self-inductance is calculated by dividing the induced emf by the rate of change of current in the coil.
Q5
The current i flowing in a coil varies with time as shown in the figure. The variation of induced emf with time would be
A primary graph showing current variation with time and four optional graphs showing the resulting induced emf.
  • A. A
    A
    Correct
  • B. B
    B
  • C. C
    C
  • D. D
    D
Explanation. Faraday's law states that induced emf depends on the rate of change of current, represented by the slope of the current-time graph.
Q6
A circular coil with a cross-sectional area of 4 cm\(^2\) has 10 turns. It is placed at the centre of a long solenoid that has 15 turns/cm and a cross-sectional area of 10 cm\(^2\). The axis of the coil coincides with the axis of the solenoid. What is their mutual inductance?
  • A. 7.54 \(\mu\)HCorrect
  • B. 8.54 \(\mu\)H
  • C. 9.54 \(\mu\)H
  • D. 10.54 \(\mu\)H
Explanation. Mutual inductance is calculated using the formula involving the number of turns, area of the inner coil, and turn density of the solenoid.
Q7
In a transformer, the number of turns in the primary and the secondary are 410 and 1230 respectively. If the current in primary is 6A, then that in the secondary coil is
  • A. 2 ACorrect
  • B. 18 A
  • C. 12 A
  • D. 1 A
Explanation. The transformer turn ratio relates the primary and secondary currents inversely; solving the ratio gives the secondary current.
Q8
A step-down transformer reduces the supply voltage from 220 V to 11 V and increase the current from 6 A to 100 A. Then its efficiency is
  • A. 1.2
  • B. 0.83Correct
  • C. 0.12
  • D. 0.9
Explanation. Efficiency is the ratio of output power to input power, calculated by multiplying voltage and current for each side.
Q9
In an electrical circuit, R, L, C and AC voltage source are all connected in series. When L is removed from the circuit, the phase difference between the voltage and current in the circuit is \(\pi/3\). Instead, if C is removed from the circuit, the phase difference is again \(\pi/3\). The power factor of the circuit is
  • A. 1/2
  • B. \(1/\sqrt{2}\)
  • C. 1Correct
  • D. \(\sqrt{3}/2\)
Explanation. Since the phase differences are equal when either component is removed, inductive and capacitive reactances are equal, indicating resonance and unity power factor.
Q10
In a series RL circuit, the resistance and inductive reactance are the same. Then the phase difference between the voltage and current in the circuit is
  • A. \(\pi/4\)Correct
  • B. \(\pi/2\)
  • C. \(\pi/6\)
  • D. zero
Explanation. The phase angle is found using the tangent of the ratio of reactance to resistance; if they are equal, the angle is 45 degrees.
Q11
In a series resonant RLC circuit, the voltage across 100 \(\Omega\) resistor is 40 V. The resonant frequency \(\omega\) is 250 rad/s. If the value of C is 4 \(\mu\)F, then the voltage across L is
  • A. 600 V
  • B. 4000 V
  • C. 400 VCorrect
  • D. 1 V
Explanation. At resonance, the voltage across the inductor equals the voltage across the capacitor, calculated from current and capacitive reactance.
Q12
An inductor 20 mH, a capacitor 50 \(\mu\)F and a resistor 40 \(\Omega\) are connected in series across a source of emf V = 10 sin 340 t. The power loss in AC circuit is
  • A. 0.76 W
  • B. 0.89 W
  • C. 0.46 WCorrect
  • D. 0.67 W
Explanation. Power loss is calculated using the rms current and resistance, where current is determined by the circuit's total impedance at the given frequency.
Q13
The instantaneous values of alternating current and voltage in a circuit are \(i = \frac{1}{\sqrt{2}} \sin(100\pi t)\) A and \(v = \frac{1}{\sqrt{2}} \sin(100\pi t + \pi/3)\) V. The average power in watts consumed in the circuit is
  • A. 1/4
  • B. 3/4
  • C. 1/2
  • D. 1/8Correct
Explanation. Average power is product of RMS voltage, RMS current, and the cosine of the phase angle between them.
Q14
In an oscillating LC circuit, the maximum charge on the capacitor is Q. The charge on the capacitor when the energy is stored equally between the electric and magnetic fields is
  • A. \(Q/2\)
  • B. \(Q/\sqrt{3}\)
  • C. \(Q/\sqrt{2}\)Correct
  • D. Q
Explanation. When energy is shared equally, the electrical energy is half the maximum energy, relating the instantaneous charge to the maximum charge by a factor of root two.
Q15
\(\frac{20}{\pi^2}\) H inductor is connected to a capacitor of capacitance C. The value of C in order to impart maximum power at 50 Hz is
  • A. 50 \(\mu\)F
  • B. 0.5 \(\mu\)F
  • C. 500 \(\mu\)F
  • D. 5 \(\mu\)FCorrect
Explanation. Maximum power transfer occurs at electrical resonance; the capacitance is determined by the resonance frequency formula given inductance and frequency.
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About these Electromagnetic Induction And Alternating Current questions

These are the Book Back multiple-choice questions for Electromagnetic Induction And Alternating Current from the Tamil Nadu State Board (Samacheer Kalvi) 12th Standard Physics syllabus. Each question shows the correct option and an original, step-by-step explanation so you understand the method, not just the answer. Use the answer key above to jump to any question, then take the practice test to check yourself under exam-like conditions.

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How many MCQs are there in Electromagnetic Induction And Alternating Current?

This chapter has 15 book-back multiple-choice questions, each with the correct answer and a step-by-step explanation.

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Yes. Every question, answer and explanation here is free, and you can also take them as a timed practice test.

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