InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 101. |
A pair of adjacent coils has a mutual inductance of 1.5 H. If the current in one coil changes from 0 to 10 A in 0.2 s, what is the change of flux linkage with the other coil? |
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Answer» Data: M = 1.5 H, I1i = 0, I1f = 10A, ∆f = 0.2s The flux linked per unit turn with the second coil due to current I1 in the first coil is Φ21 = MI1 Therefore, the change in the flux due to change in I1 is ∆21 =M(∆I1 ) = M(I1f – I1i) = 1.5 (10 – 0) = 15 Wb [Note: The rate of change of flux linkage is M(∆I1 /∆t) = 15/0.2 = 75 Wb/s] . |
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| 102. |
Why is spark produced when we break an electric circuit? |
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Answer» Due to large amount of induced e.m.f. |
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| 103. |
Consider a metal ring kept on top of a fixed solenoid (say on a carboard) (Fig 6.5). The centre of the ring coincides with the axis of the solenoid. If the current is suddenly switched on, the metal ring jumps up. Explain |
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Answer» No flux was passing through the metal ring initially. When the current is switched on, flux passes through the ring. According to Lenz’s law this increase will be resisted and this can happen if the ring moves away from the solenoid. One can analyse this in more detail (Fig 6.5). If the current in the solenoid is as shown, the flux (downward) increases and this will cause a counterclockwise current (as seen form the top in the ring). As the flow of current is in the opposite direction to that in the solenoid, they will repel each other and the ring will move upward. |
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| 104. |
Consider a metal ring kept (supported by a cardboard) on top of a fixed solenoid carrying a current I (in figure). The centre of the ring coincides with the axis of the solenoid. If the current in the solenoid is switched off, what will happen to the ring? |
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Answer» When the current in the solenoid decreases a current flows in the same direction in the metal ring as in the solenoid. Thus there will be a downward force. This means the ring will remain on the cardboard. The upward reaction of the cardboard on the ring will increase. |
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| 105. |
A metallic conductor AB moves across a magnetic field as shown in the following figure.Which of the following statements is correct? (A) The free electrons experience a magnetic force and move to the lower part of the conductor. (B) The free electrons experience a magnetic force and move to the upper part of the conductor.(C) The positive and negative charges experience a magnetic force and move, respectively, to the upper and lower parts of the conductor. (D) The moving conductor gives rise to an emf but there is no separation of charges as they are bound in the solid structure. |
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Answer» (A) The free electrons experience a magnetic force and move to the lower part of the conductor. |
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| 106. |
A metallic metre stick moves with a velocity of 2m/s in a direction perpendicular to its length and perpendicular to a uniform magnetic field of magnitude 0.2T. Find the emf induced between the ends of the stick. |
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Answer» ℓ = 1m, B = 0.2T, v = 2m/s, e = Bℓv = 0.2 × 1 × 2 = 0.4V |
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| 107. |
What are LC oscillation? and explain the generation of LC oscillation. |
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Answer» Whenever energy is given to a circuit containing a pure inductor of inductance L and a capacitor of capacitance C, the energy oscillates back and forth between the magnetic field of the inductor and the electric field of the capacitor. Thus the electrical oscillations of definite frequency are generated. These oscillations are called LC oscillations. Generation of LC oscillations: Let us assume that the capacitor is fully charged with maximum charge Q at the initial stage. So that the energy stored in the capacitor is maximum and is given by UEM = \(\frac{Q^2_m}{2C}\) As there is no current in the inductor, the energy stored in it is zero i.e., UB = 0. Therefore, the total energy is wholly electrical. The capacitor now begins to discharge through the inductor that establishes current i in clockwise direction. This current produces a magnetic field around the inductor and the energy stored in the inductor is given by UB = \(\frac{Li_2}{2}\). As the charge in the capacitor decreases, the energy stored in it also decreases and is given by UE = \(\frac{q_2}{2C}\) . Thus there is a transfer of some part of energy from the capacitor to the inductor. At that instant, the total energy is the sum of electrical and magnetic energies. When the charges in the capacitor are exhausted, its energy becomes zero i.e., UE = 0. The energy is fully transferred to the magnetic field of the inductor and its energy is maximum. This maximum energy is given by UB = \(\frac{LI^2_m}{2}\) where Im is the maximum current flowing in the circuit. The total energy is wholly magnetic. Even though the charge in the capacitor is zero, the current will continue to flow in the same direction because the inductor will not allow it to stop immediately. The current is made to flow with decreasing magnitude by the collapsing magnetic field of the inductor. As a result of this, the capacitor begins to charge in the opposite direction. A part of the energy is transferred from the inductor back to the capacitor. The total energy is the sum of the electrical and magnetic energies. When the current in the circuit reduces to zero, the capacitor becomes frilly charged in the opposite direction. The energy stored in the capacitor becomes maximum. Since the current is zero, the energy stored in the inductor is zero. The total energy is wholly electrical. The state of the circuit is similar to the initial state but the difference is that the capacitor is charged in opposite direction. The capacitor then starts to discharge through the inductor with anti-clockwise current. The total energy is the sum of the electrical and magnetic energies. As already explained, the processes are repeated in opposite direction. Finally, the circuit returns to the initial state. Thus, when the circuit goes through these stages, an alternating current flows in the circuit. As this process is repeated again and again, the electrical oscillations of definite frequency are generated. These are known as LC oscillations. In the ideal LC circuit, there is no loss of energy. Therefore, the oscillations will continue indefinitely. Such oscillations are called undamped oscillations. |
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| 108. |
What is the role of an inductor in an ac circuit ? |
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Answer» As a circuit element, an inductor slows down changes in the current in the circuit. Thus, it provides an electrical inertia and is said to act as a ballast. In a non-inductive coil (L ≅ 0), electrical energy is converted into heat due to ohmic resistance of the coil (Joule heating). On the other hand, an inductive coil or an inductor stores part of the energy in the magnetic field of its coils when the current through it is increasing; this energy is released when the current is decreasing. Thus, an inductor limits an alternating current more efficiently than a noninductive coil or a pure resistor. |
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| 109. |
Does emf rise instantaneously after the battery connected to it is switched on? |
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Answer» No. Because the back emf produced opposes the growth of current through the coil. |
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| 110. |
What is mutual induction? |
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Answer» Mutual induction is the phenomena of production of emf induced in a coil due to a change in current in a nearby coil. |
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| 111. |
Can a thin wire act as an inductor? |
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Answer» No. Because a thin wire does not enclose a significant magnetic flux. |
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| 112. |
What is the SI unit of inductance? Define it. |
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Answer» Henry (H). One henry is defined as the inductance of a coil for which there is a magnetic flux of 1 Wb is linked with it when a current of 1 A is causing it. |
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| 113. |
On what factors does the inductance of a coil depend? |
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Answer» The inductance of coil depends on the geometry of the coil and intrinsic material properties. |
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| 114. |
On what factors does the coefficient of self – inductance of a coil depend? |
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Answer» (a) The length of the solenoid (α l), (b) the number of turns per unit length in the solenoid (α n2) (c) the area of the coil (α A) (d) the permeability of the medium in ide the len id (α μr) |
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| 115. |
Define electromagnetic induction. |
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Answer» The phenomenon in which electric current is generated by varying magnetic fields is appropriately called electromagnetic induction. |
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| 116. |
State the expressions for the effective or equivalent inductance of a combination of a number of inductors connected (a) in series (b) in parallel. Assume that their mutual inductance can be ignored. |
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Answer» We assume that the inductors are so far apart that their mutual inductance is negligible. (a) For a series combination of a number of inductors, L1 , L2 , L3 , …, the equivalent inductance is Lseries = L1 + L2 + L3 + …….. (b) For a parallel combination of a number of inductors, L1 , L2 , L3 , …, the equivalent inductance is \(\cfrac{1}{L_{pa1rallel}}\) = \(\cfrac{1}{L_1}\) + \(\cfrac{1}{L_2}\) + \(\cfrac{1}{L_3}\) + … |
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| 117. |
The motion of copper plate is damped when it is allowed to oscillate between the two poles of a magnet. What is the cause of this damping? |
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Answer» As the plate oscillate, the changing magnetic flux through the plate produces a strong eddy current in the direction, which opposes the cause. Also, copper being diamagnetic substance, it gets magnetised in the opposite direction, so the plate motion gets damped. |
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| 118. |
The magnetic field within an air-cored solenoid is 0.8 T. If the solenoid is 40 cm long and 2 cm in diameter, the energy stored in its magnetic field is (A) 32 J (B) 3.2 J (C) 6.4 kJ (D) 64 kJ |
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Answer» Correct option is (A) 32 J |
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| 119. |
A simple generator has a 300 loop square coil of side 20 cm turning in a field of 0.7 T. How fast must it turn to produce a peak output of 210 V ? (A) 25 rps (B) 4 rps (C) 2.5 rps(D) 0.4 rps |
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Answer» Correct option is (B) 4 rps |
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| 120. |
What is an ac generator? State the principle of an ac generator. |
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Answer» An electric generator or dynamo converts mechanical energy into electric energy, just the opposite of what an electric motor does. Principle : An AC generator works on electromagnetic induction : When a coil of wire rotates between two poles of a permanent magnet such that the magnetic flux through the coil changes periodically with time due to a change in the angle between the area vector and the magnetic field, an alternating emf is induced in the coil causing a current to pass when the circuit is closed. |
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| 121. |
An ac generator spinning at a rate of 750 rev/min produces a maximum emf of 45 V. At what angular speed does this generator produce a maximum emf of 102 V ? |
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Answer» Data : e1 = 45 V, f1 = 750 rpm, e2 = 102 V e = NABω = NAB(2πf) ∴ e ∝ f ∴ \(\cfrac{e_2}{e_1}\) = \(\cfrac{f_2}{f_1}\) ∴ f2 = \(\cfrac{e_2}{e_1}\) x f1 = \(\cfrac{102}{45}\) x 750 = 1700 rpm This is the required frequency of the generator coil. |
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| 122. |
The current in a discharging LR circuit without the battery drops from 2.0A to 1.0A in 0.10s. (a) Find the time constant of the circuit, (b) If the inductance of the circuit is 4.0H, what is its resistance? |
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Answer» (a) For discharging circuit i = i0 e–t/τ => 1 = 2e–0.1/τ => (1/2) = e–0.1/τ => ℓn (1/2) = ℓn(e–0.1/τ) => –0.693 = –0.1/τ => τ = 0.1/0.693 = 0.144 = 0.14. (b) L = 4 H, i = L/R => 0.14 = 4/R => R = 4/0.14 = 28.57 = 28Ω. |
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| 123. |
What is electrical analogue of mass in mechanics? OR Which electrical device plays the role of electrical inertia? |
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Answer» Self – inductance. |
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| 124. |
What is back emf? |
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Answer» The emf induced in a coil which opposes the rise of current through a coil is called back emf. |
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| 125. |
Explain back emf in a motor. |
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Answer» A generator converts mechanical energy into electrical energy, whereas a motor converts electrical energy into mechanical energy. Also, motors and generators have the same construction. When the coil of a motor is rotated by the input emf, the changing magnetic flux through the coil induces an emf, consistent with Faraday’s law of induction. A motor thus acts as a generator whenever its coil rotates. According to Lenz’s law, this induced emf opposes any change, so that the input emf that powers the motor is opposed by the motor’s self-generated emf. This self-generated emf is called a back emf because it opposes the change producing it. |
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| 126. |
What is back torque in a generator? |
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Answer» In an electric generator, the mechanical rotation of the armature induces an emf in its coil. This is the output emf of the generator. Under no-load condition, there is no current although the output emf exists, and it takes little effort to rotate the armature. However, when a load current is drawn, the situation is similar to a current-carrying coil in an external magnetic field. Then, a torque is exerted, and this torque opposes the rotation. This is called back torque or counter torque. Because of the back torque, the external agent has to apply a greater torque to keep the generator running. The greater the load current, the greater is the back torque. |
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| 127. |
A motor draws more current when it starts than when it runs at its full (i.e., operating) speed. Explain. OR When a pump or refrigerator (or other large motor) starts up, lights in the same circuit dim briefly. |
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Answer» The back emf is effectively the generator output of a motor, and is proportional to the angular velocity co of the motor. Hence, when the motor is first turned on, the back emf is zero and the coil receives the full input voltage. Thus, the motor draws maximum current when it is first turned on. As the motor speeds up, the back emf grows, always opposing the driving emf, and reduces the voltage across the coil and the amount of current it draws. This explains why a motor draws more current when it first comes on, than when it runs at its normal operating speed. The effect is noticeable when a high power motor, like that of a pump, refrigerator or washing machine is first turned on. The large initial current causes the voltage at the outlets in the same circuit to drop. Due to the IR drop produced in feeder lines by the large current drawn by the motor, lights in the same circuit dim briefly. [Note : A motor is designed to run at a certain speed for a given applied voltage. A mechanical overload on the motor slows it down appreciably. If the rotation speed is reduced, the back emf will not be as high as designed for and the current will increase. At too low speed, the large current can even burn its coil. On the other hand, if there is no mechanical load on the motor, its angular velocity will increase until the back emf is nearly equal to the driving emf. Then, the motor uses only enough energy to overcome friction.] |
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| 128. |
A motor has a coil resistance of 5 Ω. If it draws 8.2 A when running at full speed and connected to a 220 V line, how large is the back emf ? |
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Answer» Data : R = 5 Ω, I = 8.2 A, eapplied = 220 V eapplied – eback =IR = 0 ∴ eback = eapplied – IR = 220 – (8.2)(5) = 220 – 42 = 178 V |
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| 129. |
A dynamo attached to a bicycle has a 200 turn coil, each of area 0.10 m2. The coil rotates half a revolution per second and is placed in a uniform magnetic field of 0.02 T. Find the maximum voltage generated in the coil. |
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Answer» Data : N = 200, A = 0.1 m2 , f = 0.5 Hz, B = 0.02T e0 = NABω = NAB (2πf) Therefore, the maximum voltage generated, e0 = (200)(0.1)(0.02)(2 × 3.142 × 0.5) = 1.26 V |
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| 130. |
An ac generator has a coil of 250 turns rotating at 60 Hz in a magnetic field of 0.6/π T. What must be the area of each turn of the coil to produce a maximum emf of 180 V ? |
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Answer» Data : N = 250, f = 60 Hz, B = \(\cfrac{0.6}π\)T e0 = NABω = NAB (2πf) ∴ A = \(\cfrac{e_0}{NB2πf}\) = \(\cfrac{180}{(250)(0.6/π)(2π×60)}\) = \(\cfrac{18}{25 \times 72}\) = 10-2 m2 This must be the area of each turn of the coil. |
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| 131. |
A light metal disc on the top of an electromagnet is thrown up as the current is switched on. Why? Give reason. |
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Answer» A metal disc is placed on the top of a magnet, as the electric current flows through the coil, an induced current in the form of Eddies flows through the metal plate, the lower face attains the same polarity, and hence the metal disc is thrown up. |
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| 132. |
A long straight current carrying wire passes normally through the centre of circular loop. If the current through the wire increases, will there be an induced emf in the loop? Justify. |
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Answer» No. Justification: As the magnetic field due to current carrying wire will be in the plane of the circular loop, so magnetic flux will remain zero. Also, magnetic flux does not change with the change in current. |
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| 133. |
State the law that gives the polarity of the induced emf. |
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Answer» Lenz’s Law: The polarity of induced emf is such that it tends to produce a current which opposes the change in magnetic flux that produces it. |
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| 134. |
What is a transformer? State the principle of working of a transformer. |
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Answer» A transformer is an electrical device which uses mutual induction to transform electrical power at one alternating voltage into electrical power at another alternating voltage (usually different), without change of frequency of the voltage. Principle : A transformer works on the principle that a changing current through one coil creates a changing magnetic flux through an adjacent coil which in turn induces an emf and a current in the second coil. |
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| 135. |
When is the flux through a surface (a) maximum? (b) zero? |
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Answer» (a) when the plane of the surface is perpendicular to the magnetic field (θ = 0º) (b) when the plane of the surface is kept parallel to the magnetic field (θ = 90º) |
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| 136. |
What is the value of the magnetic flux through a closed surface ? |
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Answer» The value of the magnetic flux through a closed surface is Zero. |
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| 137. |
Does a magnet kept near a coil induce current in it? |
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Answer» No. EMF is induced in the coil only when the magnet is moving relative the coil. |
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| 138. |
Can you induce an emf in an open circuit by electromagnetic induction? |
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Answer» Yes, induce an emf in an open circuit by electromagnetic induction. |
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| 139. |
Induced emf of electromagnetic induction depends upon:(a) Resistance on conductor(b) The value of magnetic field(c) The direction of conductor w.r.t. the magnetic field(d) Rate of change of flux linked |
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Answer» (d) Rate of change of flux linked |
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| 140. |
Mention the various energy losses in a transformer |
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Answer» Energy losses in a transformer: Transformers do not have any moving parts so that its efficiency is much higher than that of rotating machines like generators and motors. But there are many factors which lead to energy loss in a transformer. (i) Core loss or Iron loss: This loss takes place in transformer core. Hysteresis loss and eddy current loss are known as core loss or Iron loss. When transformer core is magnetized and demagnetized repeatedly by the alternating voltage applied across primary coil, hysteresis takes place due to which some energy is lost in the form of heat. Hysteresis loss is minimized by using steel of high silicon content in making transformer core. Alternating magnetic flux in the core induces eddy currents in it. Therefore there is energy loss due to the flow of eddy current, called eddy current loss which is minimized by using very thin laminations of transformer core. (ii) Copper loss: Transformer windings have electrical resistance. When an electric current flows through them, some amount of energy is dissipated due to Joule heating. This energy loss is called copper loss which is minimized by using wires of larger diameter. (iii) Flux leakage: Flux leakage happens when the magnetic lines of primary coil arc not completely linked with secondary coil. Energy loss due to this flux leakage is minimized by winding coils one over the other. |
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| 141. |
For an ideal transformer efficiency η is …… |
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Answer» For an ideal transformer efficiency η is 1 |
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| 142. |
Fig. shows two positions of a loop PQR in a perpendicular uniform magnetic field. In which position of the coil is there an induced emf? |
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Answer» In position (i), the coil remains as such in magnetic field, so there is no magnetic flux change in the coil, hence no emf is induced. In position (ii) the coil is coming out of the magnetic field, so the magnetic flux linked with it decreases and so an emf is induced in the coil. |
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| 143. |
In the figure given below, a bar magnet moving towards the right or left induces an emf in the coils (1) and (2). Find, giving reason, the directions of the induced currents through the resistors AB and CD when the magnet is moving (a) towards the right, and (b) towards the left. |
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Answer» (a) When magnet moves towards the right, the nearer faces of coils, 1 and 2 act as south poles, so current induced in AB is from B to A and in coil 2 from C to D. (b) When magnet moves towards left, the nearer faces of coils act as north poles, so current induced in coil, 1 will be from A to B and in coil 2 from D to C. |
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| 144. |
The flux rule is the terminology that Feynman used to refer to the law relating magnetic flux to emf. (RP Feynman, Feynman lectures on Physics, Vol II) |
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Answer» Modern applications of Faraday’s law of induction :
Faraday’s second law of electromagnetic induction is referred by some as the “flux rule”. |
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| 145. |
What does magnetic flux measure? |
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Answer» Magnetic flux through a surface is a measure of the number of lines of magnetic field lines passing through the surface. |
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| 146. |
Is magnetic flux scalar or a vector? |
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Answer» Magnetic flux scalar. |
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| 147. |
Show diagrammatically when is magnetic flux taken as (i) Positive (ii) Negative |
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Answer» If the normal N to area A is in the same direction to B, f = \(\vec{B}.\vec{A}\) is positive If the normal N is in the opposite direction to B. φ = cos 180° is negative. |
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| 148. |
Give the direction in which the induced current flows in the coil mounted on an insulating stand when a bar magnet is quickly moved along the axis of the coil from one side to the other as shown in the figure alongside. |
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Answer» Current induced in the coil flows clockwise for an observer sitting on the magnet. |
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| 149. |
What happens to the self – induction of a coil if a soft – iron rod is inserted into it? |
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Answer» Increases. Since iron has large permeability, the inductance increases. |
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| 150. |
Current passes through a coil shown from left to right. In which direction is the induced emf. if the current is (a) increasing with time (b) decreasing in time? |
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Answer» From Lent’s law, the induced emf must oppose the diange in the magnetic flux. (a) When the current mcreases to the right, so is the magnetic flux. To oppose the increasing flux to the tight. the induced emi Is to the left. i.e.. the point A is at a positive potential relative to point B. (b) When the current to the right is decreasing the induced emf acts to boost up the flux to the right and points to the tight, so that the point A is at a negative potential relative to point B. |
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