Access the latest CBSE Class 12 Physics Electromagnetic Induction And Alternating Current Worksheet Set A. We have provided free printable Class 12 Physics worksheets in PDF format, specifically designed for Chapter 6 Electromagnetic Induction. These practice sets are prepared by expert teachers following the 2025-26 syllabus and exam patterns issued by CBSE, NCERT, and KVS.
Chapter 6 Electromagnetic Induction Physics Practice Worksheet for Class 12
Students should use these Class 12 Physics chapter-wise worksheets for daily practice to improve their conceptual understanding. This detailed test papers include important questions and solutions for Chapter 6 Electromagnetic Induction, to help you prepare for school tests and final examination. Regular practice of these Class 12 Physics questions will help improve your problem-solving speed and exam accuracy for the 2026 session.
Download Class 12 Physics Chapter 6 Electromagnetic Induction Worksheet PDF
ONE MARK QUESTIONS
Question. The power factor of an ac circuit is 0.5. What is the phase difference between current and voltage in this circuit.
Answer : 600
Question. Define quality factor of resonance in series LCR circuit. What is its SI unit?
Answer : Q = ωr /2Δω
Question. An ac source of voltage V= V_0 sinωt is connected to an ideal inductor. Draw graphs of voltage V and current i versus ωt.
Answer :
Question. A bar magnet is moved in the direction indicated by the arrow between two coils PQ and CD. Predict the direction of the induced current in each coil.
Answer : P to Q in coil 1 and C to D in coil 2
TWO MARKS QUESTIONS
Question. Show that magnetic energy required to build up the current I in a coil of self-inductance L is given by 1/2 LI^2.
Answer : Derivation
Question. State the underlying principle of a transformer. How is the large scale transmission of electric energy over long distances done with the use of transformers?
Answer : Mutual induction. Step up transformer
Question. A light bulb is rated 100 W for 220V ac supply of 50Hz. Calculate
(a) the resistance of the bulb
(b) the rms current through the bulb
Answer : R= 484 Ω
Question. State Lenz’s law. A metallic rod held horizontally along east-west direction, is allowed to fall under gravity. Will there be an emf induced at its ends? Justify your answer.
Answer : Statement. Emf is induced.
THREE MARKS QUESTIONS
Question. Derive expression for self inductance of a long air-cored solenoid of length l, cross-sectional area A and having number of turns N.
Answer : derivation
Question. Describe briefly, with the help of a labeled diagram, the working of a step up transformer.
Answer : Brief explanation
Question. (a) Define mutual inductance and write its SI units.
(b) Derive an expression for the mutual inductance of two long coaxial solenoids of same length wound one over other.
Answer : (a) definition and hennery (b) derivation
Question. A wheel of 8 metallic spokes each 50cm long is rotated with a speed of 120 rev/min in a plane normal to the horizontal component of the Earth’s magnetic field. The Earth’s magnetic field at the plane is 0.4 G and the angle of dip is 60⁰. Calculate the emf induced between the axle and the rim of the wheel. How will be the emf is affected if the number of spokes were increased?
Answer : E= 1/2 Bωl^2
B_H=B cosδ
E= 1/2 B cosδ ωl^2=3.14×〖10〗^(-5) volt
Question.) Using phasor diagram for a series LCR circuit connected to an AC source of voltage
v= v_0 sinωτ, derive the relation for the current flowing in the circuit and the phase angle between the voltage across the resistor and the net voltage in the circuit.
Answer : Derivation
The phasor diagram of the RLC Series Circuit when the circuit is acting as an inductive circuit that means (VL>VC) is shown below and if (VL< VC) the circuit will behave as a capacitive circuit.
Steps to draw the Phasor Diagram of the RLC Series Circuit
• Take current I as the reference as shown in the figure above
• The voltage across the inductor L that is VL is drawn leads the current I by a 90-degree angle.
• The voltage across the capacitor c that is Vc is drawn lagging the current I by a 90 degree angle because in capacitive load the current leads the voltage by an angle of 90 degrees.
• The two vectors VL and VC are opposite to each other.
FIVE MARKS QUESTIONS
Question. (a) Draw a schematic sketch of an ac generator describing its basic elements. State briefly its working principle. Show a plot of variation of (i) magnetic flux and (ii) alternating emf versus time generated by a loop of wire rotating in a magnetic field (b) Why is choke coil needed in the use of fluorescent tubes with ac mains?
Answer : Brief explanation.
Principle of AC generator- It works on the principle of electromagnetic induction.
In an A.C. generator, mechanical energy is converted to electrical energy by virtue of electromagnetic induction.
* Rotation of rectangular coil in a magnetic field causes change in flux (Φ = NBACosωt).
Change in flux induces emf in the coil which is given by
ε= -dΦ/dt = NBAωSinωt ε ε= ε0Sinω
* Current induced in the coil I = ε/R = ε0Sinωt/R = I0Sinωt
,In case of choke coil power dissipation is very negligible
Question. (a) An ac source of voltage V= V_0 sinωt is connected to a series combination of L, C and R. Use the phasor diagram to obtain expressions for impedance of the circuit and phase angle between voltage and current. Find the condition when current will be in phase with the voltage. What is the circuit in this condition called?
(b) In a series LR circuit X_L=R and power factor of the circuit is P_1. When capacitor with capacitance C such that X_L=X_C is put in series, the power factor becomesP_2. CalculateP_1/P_2 .
Answer :
Question. (i) What is the function of a transformer. State it principle of working with the help of a diagram. Mention various energy losses in this device.
(ii) The primary coil of an ideal step up transformer has 100 turns and transformation ratio is also 100. The input voltage and power are respectively 220 V and 1100 W. Calculate
(a) Number of turns in secondary
(b) Current in primary
(c ) voltage across secondary
(d) Current in secondary
(e) Power in secondary
Answer : (i) Function: Transformer is used for converting
low alternating voltage at high current into high voltage at low current and vice-versa.
Principle: It works on the principle of mutual induction.
Various Energy loses in transformer are -
(a) Flux losses.
(b) Copper and Iron losses.
( c) Hysteresis Losses
(d) Humming Losses.
Step-Up Transformer:
Converts low voltage A.C. into high voltage A.C. It increase the voltage
22 Derive an expression for the average power consumed in a series LCR circuit connected to a.c. source in which the phase difference between the voltage and the current in the circuit is Φ.
Question. A circular coil of radius 5 cm has 500 turns of a wire. The approximate value of the coefficient of self induction of the coil will be:
a. 25 millihenry
b. 3 25 x 10−3 millihenry
c. 34 50 x 10−34 millihenry
d. 3 50 x 10−3 henry
Answer : A
Question. The average emf induced in a coil in which a current changes from 0 to 2 A in 0.05 s is 8 V. The self inductance of the coil is:
a. 0.1 H
b. 0.2 H
c. 0.4 H
d. 0.8 H
Answer : B
Question. A coil of Cu wire (radius-r, self inductance-L) is bent in two concentric turns each having radius is. r/2 The self inductance now:
a. 2L
b. L
c. 4 L
d. L/2
Answer : A
Question. A rectangular coil of 300 turns has an average area of 25 cm × 10 cm. The coil rotates with a speed of 50 cps in a uniform magnetic field of strength 4 × 10–2 T, about an axis perpendicular to the field. The peak value of the induced emf is: (in volt)
a. 3000 π
b. 300π
c. 30π
d. 3 π
Answer : C
Question. A wheel with ten metallic spokes each 0.50 m long is rotated with a speed of 120 rev/min in a place normal to the earth's magnetic field at the place. If the magnitude of the field is 0.04 G, the induced emf between the axle and the rim of the wheel is equal to:
a. 3 1.256 x 10−3 V
b. 6.28 × 10–3 V
c. 4 1.256 10−4 V
d. 6.28 × 10–6 V
Answer : D
Question. In the previous question at what time t instantaneous induced emf will be half of maximum induced emf:
a. 1/300 sec
b. 1/400 sec
c. 1/500 sec
d. 1/600 sec
Answer : D
Question. As shown in the figure a metal rod makes contact and complete the circuit. The circuit is perpendicular to the magnetic field with B = 0.15 Tesla. If the resistance is 3), force needed to move the rod as indicated with a constant speed of 2m/sec is:
a. 3.75 x 10–3 N
b. 3.75 × 10–2 N
c. 3.75×102 N N
d. 3.75×01–4 N
Answer : A
Question. A metal conductor of length 1 m rotates vertically about one of its ends at angular velocity 5 radians per second. If the horizontal component of earth's magnetic field is 0.2 × 10–4 T, then the e.m.f. developed between the two ends of the conductor is:
a. 5mV
b. 50μV
c. 5μV
d. 50mV
Answer : B
Question. In a region of uniform magnetic induction B = 10–2 Tesla, a circular coil of radius 30 cm and resistance π2 ohm is rotated about an axis which is perpendicular to the direction of B and which forms a diameter of the coil. If the coil rotates at 200 rpm the amplitude of the alternating current induced in the coil is:
a. 2 4π mA
b. 30mA
c. 6 mA
d. 200mA
Answer : C
Question. In the following circuit, the bulb will become suddenly bright if:
a. Contact is made or broken
b. Contact is made
c. Contact is broken
d. Won’t becomes bright at all
Answer : C
Question. A solenoid has 2000 turns wound over a length of 0.30 metre. The area of its cross-section is 1.2 × 10–3 m2. Around its central section, a coil of 300 turns is wound. If an initial current of 2A in the solenoid is reversed in 0.25 sec, then the emf induced in the coil is:
a. 6 x 10−4 V
b. 4 x 10−3 V
c. 6 x 10−2 V
d. 48 mV
Answer : D
Question. The mutual inductance between a primary and secondary circuit is 0.5 H. The resistance of the primary and the secondary circuits are 20 ohms and 5 ohms respectively. To generate a current of 0.4 A in the secondary, current in the primary must be changed at the rate of:
a. 4.0 A/s
b. 16.0 A/s
c. 1.6 A/s
d. 8.0 A/s
Answer : A
Question. An emf of 15 volt is applied in a circuit containing 5 henry inductance and 10 ohm resistance. The ratio of the currents at time t = ∞ and at t = 1 second is:
a. e1/ 2 / e1/ 2 − 1
b. e2 / e2 − 1
c. 1–e–1
d. e–1
Answer : B
Question. A simple electric motor has an armature resistance of one ohm and runs from a dc source of 12 volts. When unloaded it draws a current of 2 amperes. When a certain load is connected, its speed becomes one-half of its unloaded value. Then the current in ampere it draws is:
a. 7 amp
b. 6 amp
c. 2 amp
d. 4 amp
Answer : A
Question. A transformer is used to light 140 watt, 24 volt lamp from 240 volts ac mains. If the current in the mains is 0.7 A, then the efficiency of transformer is:
a. 63.8%
b. 84%
c. 83.3%
d. 48%
Answer : C
Question. Three inductances are connected as shown below.
Assuming no coupling, the resultant inductance will be:
a. 0.25 H
b. 0.75 H
c. 0.01 H
d. 1 H
Answer : D
Question. In the figure magnetic energy stored in the coil is:
a. Zero
b. Infinite
c. 25 J
d. None of these
Answer : C
Question. The peak value of an alternating e.m.f. E is given by E = E0 cosω t is 10 volts and its frequency is 50Hz. At time t = 1/600 sec the instantaneous e.m.f. is:
a. 10 V
b. 5 3V
c. 5V
d. 1V
Answer : B
Question. The instantaneous values of current and potential difference in an alternating circuit are i = sinω t and E = 100 cosω t respectively. r.m.s. value of wattless current (in amp) in the circuit is:
a. 1
b. 1/√2
c. 100
d. Zero
Answer : B
Question. In an ac dynamo, the number of turns in the armature are made four times and the angular velocity 9 times, then the peak value of induced emf will become
a. 36 times
b. 12 times
c. 6 times
d. 18 times
Answer : A
Question. Primary voltage is , Vp resistance of the primary winding is . Rp Turns in primary and secondary are respectively Np and Ns then secondary current in terms of primary voltage and secondary voltage respectively will be:
Answer : A
Question. In the circuit shown in the figure the ac source gives a voltage V = 20cos(2000t). Neglecting source resistance, the voltmeter and ammeter reading will be:
a. 0V,0.47A
b. 1.68V,0.47A
c. 0V,1.4A
d. 5.6V,1.4A
Answer : B
Question. What will be the equation of ac of frequency 75 Hz if its r.m.s. value is 20 A:
a. i = 20sin150πt
b. i = 20√2 sin(150πt)
c. i = 20/√2 sin(150πt)
d. i = 20√2 sin(75πt)
Answer : B
Question. At what time (From zero) the alternating voltage becomes 1/√2 times of its peak value. Where T is the periodic time:
a. T/2 sec
b. T/4 sec
c. T/8 sec
d. T/12 sec
Answer : C
Question. A 20 volts ac is applied to a circuit consisting of a resistance and a coil with negligible resistance. If the voltage across the resistance is 12 V, the voltage across the coil is:
a. 16 volts
b. 10 volts
c. 8 volts
d. 6 volts
Answer : A
Question. The potential difference V and the current i flowing through an instrument in an ac circuit of frequency f are given by 5cos V t ω = volts and I = 2 sin ωt amperes (where ω = 2π f ). The power dissipated in the instrument is:
a. Zero
b. 10 W
c. 5 W
d. 2.5 W
Answer : A
Question. If an ac main supply is given to be 220 V. What would be the average e.m.f. during a positive half cycle?
a. 198V
b. 386V
c. 256V
d. None of these
Answer : A
Question. In a series circuit R = 300Ω, L = 0.9 H , C = 2.0μ F and ω =1000rad / sec. The impedance of the circuit is:
a. 1300 Ω
b. 900 Ω
c. 500 Ω
d. 400 Ω
Answer : C
Question. In a series resonant LCR circuit, if L is increased by 25% and C is decreased by 20%, then the resonant frequency will:
a. Increase by 10%
b. Decrease by 10%
c. Remain unchanged
d. Increase by 2.5 %
Answer : C
Please click on below link to download CBSE Class 12 Physics Electromagnetic Induction And Alternating Current Worksheet Set A
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Important Practice Resources for Class 12 Physics
Chapter 6 Electromagnetic Induction CBSE Class 12 Physics Worksheet
Students can use the Chapter 6 Electromagnetic Induction practice sheet provided above to prepare for their upcoming school tests. This solved questions and answers follow the latest CBSE syllabus for Class 12 Physics. You can easily download the PDF format and solve these questions every day to improve your marks. Our expert teachers have made these from the most important topics that are always asked in your exams to help you get more marks in exams.
NCERT Based Questions and Solutions for Chapter 6 Electromagnetic Induction
Our expert team has used the official NCERT book for Class 12 Physics to create this practice material for students. After solving the questions our teachers have also suggested to study the NCERT solutions which will help you to understand the best way to solve problems in Physics. You can get all this study material for free on studiestoday.com.
Extra Practice for Physics
To get the best results in Class 12, students should try the Physics MCQ Test for this chapter. We have also provided printable assignments for Class 12 Physics on our website. Regular practice will help you feel more confident and get higher marks in CBSE examinations.
You can download the teacher-verified PDF for CBSE Class 12 Physics Electromagnetic Induction And Alternating Current Worksheet Set A from StudiesToday.com. These practice sheets for Class 12 Physics are designed as per the latest CBSE academic session.
Yes, our CBSE Class 12 Physics Electromagnetic Induction And Alternating Current Worksheet Set A includes a variety of questions like Case-based studies, Assertion-Reasoning, and MCQs as per the 50% competency-based weightage in the latest curriculum for Class 12.
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Daily practice with these Physics worksheets helps in identifying understanding gaps. It also improves question solving speed and ensures that Class 12 students get more marks in CBSE exams.
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