SOLUTION
MAGNETISM, ELECTROMAGNETIC INDUCTION & ALTERNATING CURRENT
- As you know, repulsion is the sure test for magnetism.
Correct choice: b
2. In this case, pole strength does not change but length becomes half. So moment also becomes half = 40 units
Correct choice: b
- As force is inversely proportional to square of the distance of separation, force becomes one-fourth when the distance is doubled.
Correct choice: d
- When distance is doubled, force becomes ¼, but as the pole strengths are doubled, force becomes 4 times and hence effectively force remains the same as before.
Correct choice: e
- Heating a magnetic sample results in its loosing its magnetism.
Correct choice: b
- Magnetic lines of force are closed curves, unlike electric lines of force.
Correct choice: b
- Neutral point is that at which the resultant magnetic induction is zero. So, here the magnetic field due to the magnet and that due to earth are equal and opposite.
Correct choice: d
- When a magnet is placed with its north pole facing north, its field on its axial line is towards north, i.e., in the same direction as the earth?s horizontal field. But along its equatorial line filed is directed towards south, i.e., opposite to earth?s field and hence there null point can be obtained.[magnetic field is directed away from north pole and towards south pole of a magnetic dipole]
Correct choice: b
- When placed inside an iron ring (ferromagnetic material), magnetic fields outside cannot enter the region within the ring and there is effective magnetic shielding.
Correct choice: d
- The instrument is called flux meter.
Correct choice: e
- Magnetic susceptibility is the ratio of intensity of magnetization and the magnetizing field strength, both having the same unit A/m. So it has no unit.
Correct choice: e
- Since the three magnets form a closed chain, the resultant moment is zero.
Correct choice: e
- Field due to a short dipole is inversely proportional to cube of the distance.
Correct choice: e
- Since the field due to a magnet is inversely proportional to cube of the distance of the point from the center of the dipole, and hence the fields must be in the ratio 8:1. But here the distances are not from the center and hence the ratio is not exactly 8:1.
Correct choice: c
- For getting neutral point the condition is magnet?s field = earth?s horizontal field. So here, (?0/4?) 2 M/r³ = 5 × 10 ? 5. So, r³ = 10 ? 7 × 2 × 6.75 / 5 × 10 ? 5 = 0.027. So, r = 0.3 m [?0/4? = 10 ? 7]
Correct choice: c
- Torque = M B Sin ?.
Correct choice: a
- Torque ? = M B Sin ?. So, d?/d? = M B Cos ?, which is maximum when ? = 0º
Correct choice: a
- Potential energy = - M B Cos ?. And hence the work done in turning through an angle 90º (i.e., from 0 to 90) = M B and that through 60º is = M B (1 ? cos 60) = M B (1 ? ½) = ½ MB. So, the required ratio = 2 or n = 2.
Correct choice: b
- See solution 60 first. When broken into two halves, each half has half the mass and half the length. So moment is half and moment of inertia is 1/8 times. Look at an earlier similar problem, and you will see that period becomes half the initial value = 2 s.
Correct choice: b
- Magnetic south pole, as you know, is near the geographic north.
Correct choice: a
- It is around 11.5º
Correct choice: e
- Earth?s field is totally horizontal at the equator and vertical at the poles. It is inclined to the horizontal everywhere else.
Correct choice: a
- A magnetic needle aligns itself in the direction of earth?s magnetic field, which is usually inclined.
Correct choice: b
- Dip is zero at the equator
Correct choice: a
- Cos ? = B horizontal / total B. So, here, ?3/2 = 0.5/B. Hence B = 1/?3.
Correct choice: d
- Such a line is called an aclinic line.
Correct choice: c
- Diamagnetic substances tend to move away from magnetic fields.
Correct choice: a
- It?s Curie temperature.
Correct choice: b
- Soft iron has low hysteresis loss and high permeability. It has also high susceptibility and that is why they are used to make electro-magnets.
Correct choice: b
- It gives loss of energy per cycle of magnetization and demagnetization.
Correct choice: a
- Direction of induced current, as you know. As per the law induced current will oppose the cause, which produces it.
Correct choice: c
- In accordance with Lenz?s law, the induced current in this case is such that a north pole is produced at the face that is being approached by the north pole of the magnet.
Correct choice: a
- Inductances can be added as you add resistances.
Correct choice: c
- As you know, in this case the induced e. m. f. is an a. c. and hence it changes direction every half rotation.
Correct choice: b
- Self-inductance L = ? N² A/L.
Correct choice: c
- CR has dimensions of time (called time constant of R ? C circuit) and hence 1/RC represents frequency.
Correct choice: a
- Here as there is a cut in the ring, the induced e. m. f. does not send current through the ring and hence the ring does not become a magnetic dipole and hence the magnet moves downwards undisturbed, with acceleration = g.
Correct choice: a
- Time constant of inductance is L/R
Correct choice: b
- Since there is a condenser in the circuit there cannot be any steady current, except for the charging current in the beginning.
Correct choice: d
- Inductance cannot affect steady current expect that at the time of switching on there is an opposition. So steady state current = V/R = 100/10 = 10 A.
Correct choice: a
- Frequency of an L ? C circuit is f = 1/2??LC. So when L and C are both doubled, frequency becomes half the initial value.
Correct choice: d
- Effective value or r. m. s. value of a. c. voltage = peak value/?2 = 423/1.41 = 299 V.
Correct choice: b
- Here the inductance is an ideal one and hence current lags behind the voltage by 90º
Correct choice: c
- Reactance of a capacitance = 1/C?. When frequency ?f? is doubled, ? = 2?f also doubles and hence reactance becomes half the initial value = 2.5 ohm.
Correct choice: c
- Instantaneous voltage V = V0 Sin ?t = 120?2 Sin 120 ?t = 170 Sin 120 ?/720 = 170 Sin ?/6 = 170/2 = 85 V.
Correct choice: a
- Current becomes maximum at resonance frequency ? = 1/?LC = 1/?(0.5 ×0.0008) = 1/?0.0004 = 1/0.002 = 500
Correct choice: a
- Current I = P/V = 50/100 = 0.5 A. Voltage across capacitance = ?(200² - 100²) = 100?3 = 172 V. So, reactance of capacitance (=1/C?) times current = 172. Or, (1/C 100?) 0.5 = 172. Or, C = 1/(100 ? × 344) = 9.2 × 10 ? 6 F
Correct choice: not there (such a problem is not worth attempting)
- While in series Z = ?(R² + 1/C²?²) and it is = R/?1 + ?²C²R². Given, ?(R² + 1/C²?²) =
2 R/?1 + ?²C²R². Or, (R² C² ?² + 1)/ C²?² = 4R²/(1 + ?²C²R²). Or (1 + ?²C²R²)² = 4R² C²?². So, (1 + ?²C²R²) = 2 RC?, or, (1 ? RC?)² = 0 and hence ? = 1/RC, or f = 1/2?RC
Correct choice: b *** This problem is not worth attempting.
- Voltage across L and C are always in opposite directions, or differ in phase by 180º.
Correct choice: c
- Given ? = 1/?LC, which is resonant angular frequency. So the current is maximum.
Correct choice: a
- At resonance current = 6 A. So, impedance ?Z? at resonance = R = 24/6 = 4 ohm. Hence when the same circuit is connected to a D. C of 12 V with r = 4 ohm, current = V/(R + r) = 12/8 = 1.5 A.
Correct choice: c
- At resonance impedance = resistance = 20 ohm.
Correct choice: a
- Since voltage across inductance and capacitance is the same, it is implied that the circuit resonates with the applied a. c. So impedance is resistance. Hence when resistance is short circuited, R becomes zero and hence current becomes infinity.
Correct choice: b
- It is admittance.
Correct choice: d
- At resonance frequency p. d. across inductance = p. d. across capacitance and opposite. So the total p. d. across inductance and capacitance together = 0, and hence A reads zero.
Correct choice: a
- In this case a north pole has to be produced and hence the current induced is in the anticlockwise direction.
Correct choice: b
- Tangent of dip = vertical component / horizontal component = 1/?3. So dip = 30º
Correct choice: a
- There can be magnets without poles and with three poles.
Correct choice: d
- Work done in this case = 2 M B = 2 times 2.5 times 0.2 = 1 J
Correct choice: b
- Period of a magnet in a magnetic field T = 2? ?(I / M B), where I is moment of inertia = mass times length square divided by 12. Here in the first case, I = I and when cut into two pieces lengthwise, mass is half but length is the same, and hence moment of inertia becomes half. Also, moment is half the initial value. Therefore there is no change in period.
Correct choice: a
- In this case it can be shown that the force is inversely proportional to fourth power of distance.
Correct choice: d
62. Here again the deduction of the result is a bit lengthy. Now, you simply accept that the result is ?a?.
Correct choice: a
63. If you look at the formula for inductance, L = ? N² A / l, we see that the unit of permeability is henry/meter, since A has unit m², length ?l? has unit m and N, number of turns has no unit.
Correct choice: c
64. This is somewhat similar to problem no. 62. True dip is 45. So tan 45 = 1 = BV / BH. And if ? is the apparent dip, then tan ? = BV / BH cos 60 = 1/(1/2) = 2. So ? is a little more than 60º [since tan60 = 1.732] and so the correct value is 63º
Correct choice: a
65. When placed with northpole facing north, field on the axial line is in the direction of earth?s horizontal field and that on the equatorial line is opposite to it. So null point will be on the equatorial line.
Correct choice: b
66. Here the resultant moment is zero as the north and south together cancel each other.
Correct choice: d
67. When the dip needle, the instrument used to measure dip, is placed with its plane perpendicular to magnetic meridian, the field acting on the needle of the device is vertical earth?s field and hence the needle will stand vertical.
Correct choice: b
68. It is the characteristic of a diamagnetic substance.
Correct choice: d
69. Induced e. m. f. = ½ B L² ? = ½ 0.1 × 0.15² × 60 = 0.0675 V = 67.5 m V.
Correct choice: a
70. Since the current through the coils is in the same direction, the field between the coils is the sum of the fields produced by both of them. So, as they are brought closer, field strength increases and hence the induced current will be such as to reduce the field. In other words current in the coils decrease.
Correct choice: b
71. Current in the larger loop I = Voltage / resistance = (4 + 2.5 t)/(2? × 1 × 0.1 m ohm) = 6.37 + 3.98 t (m A). So, field on the axis of the larger coil = field on the axis of the smaller coil B = ?0 I/ 2 a. Put the values of the quantities and get B. Now, flux through the smaller loop ? = B A. Induced e. m. f. in it = d?/dt. Induced current = induced e. m. f. /resistance, 2?×0.1×0.1×10 ? 3. This will work out to 1.25 A. *** This is a problem, not worth attempting for our entrance examination.
Correct choice: b
72. Vertical component of earth?s field = horizontal component times tangent of dip = 0.02 tan 60 = 0.035 m T. So the induced e. m. f. = B L v = 0.035 millitesla times 20 times 250 = 0.175 V
Correct choice: d
73. When terminal velocity is reached, net force = 0. Or, weight of the rod = magnetic force. Or, 0.2 times 10 = B I L = 0.6 times I times 1. SO, I = 2/0.6 = 3.33 A
Correct choice: b
74. When the magnet is moved towards the loop, induced current in the loop will tend to oppose the relative motion and hence the coil moves backwards.
Correct choice: a
75. Time constant is time taken by current to become 63.2% of maximum.
Correct choice: e
76. Here the inductance becomes 2 times the initial value, since L depends directly on square of N and inversely on length.
Correct choice: a
77. Self-induced e. m. f. = L dI/dt. So, 2 = L (6/0.03) = 200 L. So, L = 0.01 H
Correct choice: d
78. Induced e. m. f. is directly proportional to L and hence the answer here is 8/2 = 4.
Correct choice: a
79. The quantity Planck?s constant ?h?/electronic charge ?e? has the same dimensions as magnetic flux.
Correct choice: c
80. Energy stored = ½ L I². So, when the current is doubled, energy becomes 4 times.
Correct choice: b
81. When the rate of change of current is steady, induced current will also be steady, but always opposing the inducing current.
Correct choice: b
82. When turned through 180º, flux through the coil changes from + ? to ? ?, and so the change is 2? = 2 B A N = 2 times 0.04 (m T) times 0.05 times 1000 = 0.004. So the induced e. m. f. = 0.004/0.1 = 0.04 V = 40 m V.
Correct choice: c
83. In this case the induced e. m. f = B L v = 0.02 milli times 1 times 20 = 0.4 m V
Correct choice: a
84. Look at the formula in solution no. 77, you can see that the unit of L is V s / A.
Correct choice: c
85. During switching on and off the inductance exhibits self-induction, while the resistance does not.
Correct choice: c
86. Here induced voltage = L dI/dt = 0.5 milli times 1.5 / 2.5 micro = 0.3 kilo volt = 300 V
Correct choice: e
87. So as to reduce eddy current as laminated core has high resistance.
Correct choice: d
88. As copper plate is pushed into the space between the poles, the induced eddy current will oppose this motion and hence the face of the plate nearer to the north pole, will attain north polarity, which means the current is in the anticlockwise direction.
Correct choice: b
89. From the equation we see that ? = 2? f = 377. So, f = 60 hz.
Correct choice: a
90. R. M. S. value of the given a. c. is = 10/?2 = 5?2. So the resultant of the a. c. and D. C. currents = ?[5² + (5?2)²= ?75 = 5?3 A.
Correct choice: e
91. Reactance = L ?. So, here 100 = L × 2?f. So, L = 1/? = 0.32 H, since frequency of our domestic a. c. supply = 50 Hz.
Correct choice: a
92. Peak current = peak voltage / L ? = 10/100 times 5 mH = 10/0.5 = 20 A
Correct choice: d
93. Battery provides D. C. which is unaffected by inductance and hence current = V/R = 10/10 = 1 A.
Correct choice: a
94. Hot wire ammeter can work on both a. c. and D. C, since heat is proportional to I².
Correct choice: a
95. D. C is 2 A and effective value of a. c. is 2/?2 = ?2 A. As heat is proportional to I², it is in the ratio 4:2 = 2:1
Correct choice: d
96. At steady state the voltage across the coil of the motor = 20 times 1.5 = 30 V. So, 200 ? 30 = 170 V is the back e. m. f.
Correct choice: b
97. Resonant frequency depends on L and C and not on R. So change in R will not affect it.
Correct choice: e
98. When an a. c. is passed through the wire the alternating magnetic field interacts with the magnet?s field and hence the string is set into vibration.
Correct choice: e
99. Peak current = peak voltage/resistance = 200?2/400 = 1/?2 A = 0.71 A
Correct choice: b
100. Induced current = induced voltage/resistance = B A N ?/r = 0.01 times pi times 0.3 square times 5 times 2 pi times 200/60 divided by 9.87 = 0.03 A = 30 m A.
Correct choice: a
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