Multiple Choice Questions Involving Electromagnetic Induction

“There is only one corner of the universe that you can be certain of improving, and that’s your own self.”

– Aldous Huxley

Many questions on electromagnetic induction have been discussed on this site earlier. You can access those questions by clicking on the label, ‘electromagnetic induction’ below this post. Today we shall discuss a few more questions in this section.

(1) An object of mass M (Fig.) carrying a charge +Q is released from rest at the top of a smooth inclined plane of inclination θ. A magnetic field of flux density B acts  throughout the region and is directed normally into the plane of the figure. Which one among the following statements is correct?

(a) The acceleration of the object down the incline depends on Q but does not depend on B.

(b) The acceleration of  the object down the incline depends on B but does not depend on Q.

(c) The acceleration of the object down the incline depends on both Q and B.

(d) The acceleration of the object down the incline does not depend on Q and B.

(e) All the above statements are incorrect.

The component of the gravitational force along the incline is Mgsinθ which can produce in the object an acceleration gsinθ down the incline. When the object moves down along the inclined plane, the magnetic force (Lorentz force) acting on the object is directed perpendicular to the inclined plane such that it tries to press the object against the inclined plane.

[Apply Fleming’s left hand rule to arrive at the direction of the magnetic Lorentz force].

Since the magnetic force on the object is perpendicular to the inclined plane, it has no component along the plane. The acceleration of the object down the incline is gsinθ itself and is independent of Q and B [Option (d)].

(2) In the above question suppose the inclined plane is rough and the object is able to slide down. Which one among the following statements is correct?

(a) The acceleration of the object down the incline depends on Q but does not depend on B.

(b) The acceleration of the object down the incline depends on B but does not depend on Q.

(c) The acceleration of the object down the incline depends on both Q and B.

(d) The acceleration of the object down the incline does not depend on Q and B.

(e) All the above statements are incorrect.

In this case the frictional force acting between the object and the inclined plane reduces the acceleration of the object. The frictional force F depends on the normal force N acting on the object.

[Remember that F = μN where μ is the coefficient of friction]

But the normal force (which is Mgcosθ + magnetic Lorentz force) depends on both Q and B since the magnetic Lorentz force is QvB where v is the velocity of the object.

Therefore, the acceleration of the object down the incline depends on both Q and B [Option (c)].

(3) A plane rectangular coil of length 2L and breadth L has N turns of insulated copper wire in it. Initially (at time t = 0) the plane of the coil is perpendicular to a uniform magnetic field of flux density B . The coil rotates with angular velocity ω such that the axis of rotation of the coil is at right angles to the magnetic field (Fig.). If the voltage induced in the coil is V_maxsinωt, the angular velocity of the coil is

(a) V_max / 2NL²B

(b) 2V_max t / NL²B

(c)V_max/NL²B                                                                                            

(d) NL²B/ V_max

(e) 2NL²B/ V_max

The magnetic flux φ linked with the coil at the instant t is given by      

φ = NAB cos ωt where A is the area of the coil.

The voltage V induced in the coil is is given by

            V = – dφ /dt = NABω sin ωt

It is given in the question that V = V_maxsinωt

Therefore, V_max = NABω

This gives ω = V_max /NAB

Since the area A = 2L², we have ω = V_max / 2NL²B

(4) In the transformer shown in the adjoining figure, the secondary winding has taps A, B, C, D and E. The number of turns between taps A and B is 100. The number of turns between taps B and C is 50, the number of turns between C and D is 40 and that between D and E is 10. The primary has 2000 turns and is connected to 220 volt A. C. mains. Between which taps will you obtain 5.5 volt output? 

(a) Between A and B

(b) Between A and B as well as B and E

(c) Between B and C as well as C and E

(d) Between C and D

(e) Between D and E

Since the primary voltage is 220 V and the number of turns in the primary is 2000, the induced voltage per turn is 220/2000 = 0.11 volt.

Therefore, the number of turns required to produce a voltage of 5.5 volt is (5.5)/(0.11) = 50.

Since there are 50 turns between taps B and C as well as between taps C and E, the correct option is (c).

[You can work this out this way also:

Considering the entire secondary of the transformer, the secondary to primary turns ratio is 200/2000 = 1/10. Therefore, the secondary voltage (between the ends A and E) is 220/10 = 22 V. Since the voltage across the entire secondary containing 200 turns is 22 volt, we will get 5.5 volt across 50 turns (between taps B and C as well as between taps C and E)]

 

Kerala Engineering Entrance 2009 Questions (MCQ) on Alternating Current Circuits

Imagination is more important than knowledge
– Albert Einstein

Alternating current circuits were discussed on this site in the post dated 26th October 2006. You will find the post here.

Today we will discuss the questions from alternating currents which appeared in Kerala Engineering Entrance 2009 question paper. Here are the questions:

(1) In an LCR series ac circuit the voltage across L, C and R is 10 V each. If the inductor is short circuited, the voltage across the capacitor would become

(a) 10 V

(b) 20/√2 V

(c) 20√2 V

(d) 10/√2 V

(e) 20 V

Since the voltages across L and C are of the same value, the circuit is at resonance and Lω = 1/Cω where ω is the angular frequency. Under resonance the voltage across the resistor is equal to the supply voltage. So the supply voltage is 10 V.

When the inductor is short circuited, the LCR circuit reduces to an RC circuit and the supply voltage is the vector sum of the voltages across R and C so that we have

10 = √(V_C² + V_R²)

The voltage drops V_C and V_R across the capacitor and resistor respectively are equal in value since Lω = 1/Cω = R as judged from the equal voltage drops across L, C and R in the series LCR circuit. Therefore, from the above equation V_C = V_R = 10/√2 V

(2) A transformer of efficiency 90% draws an input power of 4 kW. An electrical appliance connected across the secondary draws a current of 6 A. The impedance of the device is

(a) 60 Ω

(b) 50 Ω

(c) 80 Ω

(d) 100 Ω

(e) 120 Ω

Efficiency, η = P_o/P_i with usual notations. Therefore, the output power is given by

P_o = η P_i = 0.9 ×4000 W = 3600 W.

In the problem nothing is mentioned about the power factor cos φ. At your level the load is usually assumed to be effectively resistive so that the power factor may be assumed to be unity. Therefore we have

P_o = I_o²R where R is the load resistance (or, the load impedence in this case).

This gives R = P_o /I_o² = 3600 /6² = 100 Ω.

(3) The impedance of an R-C circuit is Z₁ for a frequency f and Z₂ for frequency 2f. Then Z₁/Z₂ is

(a) between 1 and 2

(b) 2

(c) between ½ and 1

(d) ½

(e) 4

The impedance of an R-C circuit is √(R² + 1/C²ω²).

Since the angular frequency ω = 2πf we have

Z₁/Z₂ = [√(R² + 1/C²ω²)] /[√(R² + 1/4C²ω²)]

For large value of ω the value of Z₁/Z₂ tends to 1.

For small value of ω the value of Z₁/Z₂ tends to 2.

Therefore Z₁/Z₂ lies between 1 and 2.

(4) When a circular coil of radius 1 m and 100 turns is rotated in a horizontal uniform magnetic field, the peak value of emf induced is 100 V. The coil is unwound and then rewound into a circular coil of radius 2 m. If it is rotated now, with the same speed, under similar conditions, the new peak value of emf developed is

(a) 50 V

(b) 25 V

(c) 100 V

(d) 150 V

(e) 200 V

When a coil having N turns and area A rotates in a uniform magnetic field B with angular velocity ω, the peak value of the emf (V_max) induced in the coil is given by

(V_max) = NABω

The area of the coil in the second case is 4A since the radius is doubled. But the number of turns will be N/2 in the second case since the length of the wire required for a turn is doubled.

If V₁ and V₂ are the peak values of the emf in the two cases, we have

V₁/V₂ = 100 /V₂ = NABω / [(N/2)×(4A)Bω]

Or, 100 /V₂ = ½ from which V₂ = 200 volt.

Kerala Govt Engineering Entrance 2007-Questions on Alternating Currents

Kerala Government Engineering Entrance 2007 question paper contained four questions from alternating currents out of 72 questions in the physics part. Here are the questions:

(1) A square coil of side 25 cm having 1000 turns is rotated with uniform speed in a magnetic field about an axis perpendicular to the direction of the field. At an instant t the e.m.f. induced in the coil is e = 200 sin100πt. The magnetic induction is

(a) 0.50 T (b) ) 0.02 T (c) ) 10^–3 T (d) ) 0.1 T (e) ) 0.01 T

If you remember the production of alternating voltage when a plane coil rotates in a magnetic field, you will be able to answer this question in no time. The induced alternating e.m.f. is

e = naBω sin (ωt) and the maximum emf induced is naBω where ‘n’ is the number of turns, ‘a’ is the area (of the coil), B is the magnetic induction and ‘ω’ is the angular speed. So, we have

1000×(25×10^–2)²×B×100π = 200. [The value of ‘ω’ is obtained from the form of the equation for the emf].

This gives B = 0.01 T

(2) When a d.c. voltage of 200 V is applied to a a coil of self inductance 2√3/π H, a current of 1 A flows through it. But by replacing d.c. source with a.c. source of 200 V, the current in the coil is reduced to 0.5 A. Then the frequency of a.c. supply is

(a) 100 Hz (b) 75 Hz (c) 60 Hz (d) 30 Hz (e) 50 Hz

In an LR circuit the direct current is limited by resistance only. On applying a direct voltage of 200V, the current flowing is 1 A which means that the resistance of the coil is 200 Ω.

When the a.c. source is connected, the current is given by I = V/Z where the impedance Z is given by Z = √(R²+L² ω²)

Therefore 0.5 = 200/√[200²+(2√3/π)² ω²] = 200/√[200²+(2√3/π)² (2πf)²], from which f = 50 Hz.

(3) In a LR circuit, the value of L is (0.4/π) henry and the value of R is 30 Ω. If in the circuit, an alternating emf of 200 V at 50 cycles per second is connected, the impedance of the circuit and the current will be

(a) 11.4 Ω, 17.5 A (b) 30.7 Ω, 6.5 A (c) 40.4 Ω, 5 A

(d) 50 Ω, 4 A (e) 35 Ω, 6.5 A

This question is quite simple and straight forward.

Impedance, Z = √(R²+L² ω²) = √[(30²+(0.4/π)² (100π)²)] since ω = 2πf =2π×50 = 100π.

This gives Z = 50 Ω.

Current I = V/Z = 200/50 = 4 A.

(4) A transformer has an efficiency of 80%. It is connected to a power input of 5 kW at 200 V. If the secondary voltage is 250 V, the primary and secondary currents are respectively

(a) 25 A, 20 A (b) 20 A, 16 A (c) 25 A, 16 A (d) 40 A, 25 A (e) 40 A, 16 A

This question also is quite simple and can be worked out within a minute.

Primary current I_P = (5000 watts)/(200 volts) = 25 A.

Secondary current = (5000×0.8 watts)/(250 volts) = 16 A.

[Note that the output power from the secondary side is efficiency times the input power].

M.C.Q. on A. C. Circuits

Problems in alternating current circuits are generally interesting and if you have a thorough idea about the vector method (phasor method), most problems become quite simple. To illustrate the method, let us consider the following MCQ:

In a series LCR circuit, the voltages across the inductor and capacitor are 18V and 12V respectively. If the output of the alternating voltage source connected to this series LCR circuit is 10V, what is the voltage drop across the resistor?

(a) 6V (b) 8V (c)10V (d) 12V (e) 4V

In the vector method, alternating currents and voltages are treated as vectors. The currents and voltages used here can be peak values or RMS values. Let us take RMS values since all currents and voltages mentioned in alternating current problems are RMS values (by convention) unless specified otherwise. You should note that the current and voltage are in phase in the case of a resistor. But the voltage across an inductor leads the current by π/2 where as the voltage across a capacitor lags behind the current by π/2. This means that the voltage drops (vectors) across an inductor and a capacitor are oppositely directed. The resultant voltage drop across the series LC combination in the above question is therefore 18V-12V= 6V and this voltage vector leads the voltage (vector) across the resistor by π/2. The supply voltage vector (10V) is the vector sum of the voltage drop across the resistor and the net voltage drop across the LC combination. [All these vectors are shown in the figure in which the current in the circuit is considered as a vector I in the positive X-direction. Note that this current vector is common for all the components since the circuit is a series circuit.. The voltage drop (vector V_R) across the resistor also is taken along the positive X-direction since the current and voltage are in phase in the case of a resistor. The voltage across the inductor is indicated by the vector V_L which leads the current vector I by π/2. It is therefore shown in the positive Y-direction. The voltage across the capacitor is indicated by the vector V_C which lags behind the current vector I by π/2. It is therefore shown in the negative Y-direction.]

Therefore we have,
10 = √(V_R² + 6²) where V_R is the voltage drop across the resistor. From this we obtain V_R= 8V.
In general, the supply voltage (V) is given by
V = √[V_R² +(V_L-V_C)²]
Here is another simple question:
An inductor is connected in series with a resistor. When a 15V, 50 Hz alternating voltage source is connected across this series combination, the voltage drop across the inductor is 5V. The voltage drop across the resistor will be
(a) 15V (b) 20V (c) 10V (d) 2√10 V (e) 10√2 V
The voltage drop vector V_R in the case of the resistor is in phase with the current vector. But the voltage drop vector V_L in the case of the inductor leads the current vector by π/2. Hence we have two vectors of magnitudes V_R volts and 5 volts with an angle π/2 between them. Their vector sum gives the supply voltage vector of magnitude 15 volts. Hence we have,
15 = √ [V_R² + 5²], from which V_R = √200 = 10√2 volts.
What do you grasp from the above discussion? Certainly this: You cannot add voltage drops in AC circuits as scalar quantities. You have to perform vector addition taking into account the phase relation between the current and the voltage in the components. Here you have to forget your habit of scalar addition you perform in the case of direct current circuits!
The following MCQ appeared in the Karnataka CET (Medical) 2002 question paper:
When 100V D.C. is applied across a coil, a current of 1A flows through it. When 100V, 50Hz A.C. is applied to the same coil, only 0.5 A flows. The inductance of the coil is
(a) 5.5 mH (b) 0.55 mH (c) 55 mH (d) 0.55 H
Instead of giving you the resistance of the coil, you are given the direct voltage and the resulting direct current to enable you to calculate the resistance: R = V/I =100/1 =100 Ω. When the alternating voltage is applied, we have I = V/Z = V/√(R² + L² ω² ) so that 0.5 = 100/√[100² + L² ×(100π)²]. Squaring, 0.25 = 10⁴/ (10⁴ + 10⁴π²L²). Since π² = 10 (nearly), this reduces to 0.25 = 1/(1 + 10L²), from which L = 0.55henry = 55 mH.
Now consider the following multiple choice question:
A series LCR circuit is connected to a 10V, 1 kHz A.C. supply. If L= 40mH, R= 50Ω and the current in the circuit is 0.2A, the voltage drop across the capacitor (in volts) is approximately
(a) 10V (b) 15V (c) 20V (d) 25V (e) 50V
The value of the capacitor is not given and you are asked to find the voltage drop across it. This is a problem involving resonance, indicated by the current of 0.2A (=10/50) which is limited by the resistor only. At resonance, the voltage drop across the capacitor is equal (in magnitude) to the voltage drop across the inductor, which is ILω = 0.2X0.04 × 2π×1000 = 16π = 50V, nearly. Therefore, the correct option is (e).
Let us now discuss the following question which appeared in the AIIMS 2003 question paper:
A capacitor of capacitance 2μF is connected in the tank circuit of an oscillator oscillating with a frequency of 1 kHz. If the current flowing in the circuit is 2 mA, the voltage across the capacitor will be
(a) 0.16V (b) 0.32V (c) 79.5V (d) 159V
This is an example of how simple a question can be sometimes. The voltage drop across the capacitor = I/Cω = (2×10^-3 )/(2 ×10^-6×2π×1000) =0.16V.
The following MCQ also is a simple one, but some of you may be tempted to give the wrong answer.
In an oscillating LC circuit the maximum charge on the capacitor is Q. The charge on the capacitor when the energy is shared equally between the electric and magnetic fields is
(a) Q (b) Q/2 (c) Q/√3 (d) Q/√2
Energy in the electric field means the energy of the capacitor and the energy in the magnetic field means the energy of the inductor. The total energy output of the oscillator is the maximum energy stored in the capacitor (or inductor), which is equal to Q²/2C. When the energy is shared equally between the electric and magnetic fields, the energy of the capacitor = ½ × Q²/2C. If ‘q’ is the charge on the capacitor in this condition, we have,
q²/2C = ½ × Q²/2C, so that q = Q/√2.
To conclude this post, let us discuss one more question:
In a series LCR circuit the voltage across each of the components L,C, and R is 80V. The supply voltage and the voltage across the LC combination are respectively
(a) 80V, 160V (b) 240V, 0V (c) 160V, 0V (d) 80V, 160V (e) 80V, 0V
Since the voltage across the capacitor and the inductor are equal in magnitude, the circuit is at resonance. Since the voltages (vectors!) across these components are in opposition, the net voltage across them (LC combination) is zero. The entire supply voltage will appear across the resistor. This is equal to 80V, as given in the question. So, the correct option is (e).

Very simple questions may tempt you to be careless while answering. Don’t be careless while answering this simple question which appeared in H.P.P.M.T. 2005 question paper:
An ideal transformer has N_p turns in the primary and N_s turns in the secondary. If the voltage per turn is V_p in the primary and V_s in the secondary, V_s/V_p is equal to
(a) 1 (b) N_s/N_p (c) N_p/N_s (d) (N_p/N_s)²
You should note that the magnetic flux linked per turn is the same for the primary and the secondary, since the coupling coefficient between them is almost unity. The rate of change of flux per turn also is the same, which means the induced voltage per turn is the same for the primary and the secondary. Therefore, the correct option is (a).

The following M.C.Q. appeared in Kerala Medical Entrance 2006 test paper:
A 16 μF capacitor is charged to 20 volt potential. The battery is then disconnected and pure 40 mH coil is connected across the capacitor so that LC oscillations are set up. The maximum current in the coil is
(a) 0.2 A (b) 40 mA (c) 2 A (d) 0.4 A (e) 0.8 A
‘Pure’ 40 mH coil means the resistance of the coil is zero. So, there is no energy loss due to Joule heating and the entire initial energy of the capacitor is handed over to the inductor when the current in the circuit is maximum. Therefore we have, ½CV² = ½ L I² with usual notations so that I = √[CV²/L] = √[(16×10^-6×400)/ (40×10^-3)] = 0.4 A