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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.
| 51. |
Consider an infinitely long wire carrying a current I (t), With `(dI)/(dt) = lambda` = constant . Find the current produced in the rectangular loop of wire ABCD if its resistance is R, Fig |
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Answer» In Fig. XY is infinitley long wise carrying current I (t), such that `(dI)/(dt) = lambda =` constant. At any distance r from the wire, strength of manetic field, `B (r ) = (mu_(0) I)/(2 pu r)`. It is directed out of the pape. Total magnetic flux linked with the rectangular loop `phi = int_(x_(0))^(x) B (r ). dA = int_(x_(0))^(x) (mu_(0) I)/(2 pi r) ldr` `phi = (mu_(0) Il)/(2 pi) (log_(e)) (x)/(x_(0)) = LI`, where L is self-inductance of the loop `L = (mu_(0) l)/(2 pi) (log_(e)) (x)/(x_(0))` Induced current, `I = (e)/(R ) = (L)/(R ) (dI)/(dt) = (L)/(R ) lambda = (mu_(0) l lambda)/(2 pi R) (log_(e)) (x)/(x_(0))` |
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| 52. |
A square loop of side 10 cm and resistance 0.5 `Omega` is placed vertically in the east-west plane. A uniform magnetic field of 0.10 T is set up across the plane in the north-east direction. The magnetic field is decreased to zero in 0.70 s at a steady rate. Determine the magnitudes of induced emf and current during this time-interval. |
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Answer» The angle `theta` made by the area vector of the coil with the magnetic field is `45^(@)`. From Eq. (6.1), the initial magnetic flux is `phi=BA costheta` `=(0.1xx10^(-2))/sqrt2Wb` Final flux, `phi_(min)=0` The change in flux is brought about in 0.70 s. From Eq. (6.3), the magnitude of the induced emf is given by `epsi=(abs(Deltaphi_(B)))/(Deltat)=(abs((phi-0)))/(Deltat)=10^(-3)/(sqrt2xx0.7)=1.0 mV` And the magnitude of the current is `I=epsi/R=(10^(-3)V)/(0.5Omega)=2mA` Note that the earth’s magnetic field also produces a flux through the loop. But it is a steady field (which does not change within the time span of the experiment) and hence does not induce any emf. |
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| 53. |
A metallic rod of 1 m length is rotated with a frequency of 50 `rev//s`, with on end hinged at the centre and the other end at the circumference of a circular metallic ring of radius 1 m, about an axis passing through the centre and perpendicular at to the plane of the ring. A constant uniform magnetic field of 1 T parallel to the axis is persent eveywhere. what is the e.m.f. between the centre and the metallic ring? |
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Answer» Method I As the rod is rotated, free electrons in the rod move towards the outer end due to Lorentz force and get distributed over the ring. Thus, the resulting separation of charges produces an emf across the ends of the rod. At a certain value of emf, there is no more flow of electrons and a steady state is reached. Using Eq. (6.5), the magnitude of the emf generated across a length dr of the rod as it moves at right angles to the magnetic field is given by d d Bv r `epsi` = . Hence, `epsi=intdepsi=underset(O)overset(R)intBvdr=underset(O)overset(R)intBomegardr=(BomegaR^(2))/2` Note that we have used v = `omega` r. This gives `epsi=1/2xx1.0xx2pixx50xx(l^(2))` = 157 V Method II To calculate the emf, we can imagine a closed loop OPQ in which point O and P are connected with a resistor R and OQ is the rotating rod. The potential difference across the resistor is then equal to the induced emf and equals B `xx` (rate of change of area of loop). If `theta` is the angle between the rod and the radius of the circle at P at time t, the area of the sector OPQ is given by `piR^(2)xxtheta/(2pi)=1/2R^(2)theta` where R is the radius of the circle. Hence, the induced emf is `epsi=Bxxd/(dt)[1/2R^(2)theta]=1/2BR^(2)(d theta)/(dt)=(BomegaR^(2))/2` [Note: `(d theta)/(dt)=omega=2piv]` This expression is identical to the expression obtained by Method I and we get the same value of `epsi`. |
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| 54. |
A cylindrical bar magnet is kept along the axis of coil. Will there be a current induced in the coil. Will there be a current induced in the coil if the magnet is rotaded about its axis ? Give reasons. |
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Answer» No, because `phi = NBA = `constant `:.e = (d phi)/(dt) = 0 , i. = 0 ` |
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| 55. |
(a) A closed loop is held stationary in the magnetic field between the north and south poles of two permanent magnets held fixed. Can we hope to generate current in the loop by using very strong magnets? (b) A closed loop moves normal to the constant electric field between the plates of a large capacitor. Is a current induced in the loop (i) when it is wholly inside the region between the capacitor plates (ii) when it is partially outside the plates of the capacitor? The electric field is normal to the plane of the loop. (c) A rectangular loop and a circular loop are moving out of a uniform magnetic field region (Fig. 6.8) to a field-free region with a constant velocity v. In which loop do you expect the induced emf to be constant during the passage out of the field region? The field is normal to the loops. (d) Predict the polarity of the capacitor in the situation described by Fig. 6.9. |
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Answer» (a) No. However strong the magnet may be, current can be induced only by changing the magnetic flux through the loop. (b) No current is induced in either case. Current can not be induced by changing the electric flux. (c) The induced emf is expected to be constant only in the case of the rectangular loop. In the case of circular loop, the rate of change of area of the loop during its passage out of the field region is not constant, hence induced emf will vary accordingly (d) The polarity of plate ‘A’ will be positive with respect to plate ‘B’ in the capacitor. |
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| 56. |
A small, conducting circular loop is placed inside a long solenoid carrying a current. The plane of the loop contains the axis of the solenoid. If the current in the solenoid is varied, the current induced in the loop isA. anticlockwiseB. clockwiseC. zeroD. clockwise or anticlockwise depending on whether the resistance in increased or decreased. |
| Answer» Correct Answer - C | |
| 57. |
A small, conducting circular loop is placed inside a long solenoid carrying a current. The plane of the loop contains the axis of the solenoid. If the current in the solenoid is varied, the current induced in the loop isA. clockwiseB. anti-clockwiseC. zeroD. clockwise or anti-clockwise depending on whether the resistance is increased or decreased |
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Answer» Correct Answer - c `phi=0`. Therefore, `Deltaphi=0` |
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| 58. |
A cylindrical bar magnet is kept along the axis of circular coil and near it as shown in Fig. Will there be any induced emf at the terminals of the coil, when magnet is rotated (a) about its own axis (b) about an aixs perpendicular to the length of the magnet ? |
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Answer» (a) When the magnet is rotated about its own axis, there is no change in magnetic flux linked with the coil. Therefore, induced emf = 0. (b) When the magnet is rotated about an axis perpendicular to its length, orientation of the magnet flux linked with the coil changes. Hence an emf in induced in the coil. |
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| 59. |
A conducting circular loop is placed is a uniform magnetic field B =0.20 T with its plane perpendicualr to the field . Somehow, the radius of the loop starts shrinking at a constant rate of `1.0 mm s^(-1)`. Find the induced emf in the loop at an instant when the radius is 2 cm. |
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Answer» Here, `B = 0.02 T`, `(dr)/(dt) = 1.0 mm//s = 10^(-3) m // s` `e = ?, r = 2 cm = 2 xx 10^(-2) m` As `phi = BA = B (pi r^(2))` `:. |e| = (d phi)/(dt) = B (2 pi r) (dr)/(dt)` `0.02 xx 2 xx 3.14 xx 2 10^(-2) xx 10^(-3)` `= 2.5 xx 10^(-6) V` |
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| 60. |
When a magnet is moved towards a coil the direction of induced current is clockwise. If the magnet is moved away from the coil, the direction of induced current will beA. clockwiseB. anticlockwiseC. zeroD. any direction |
| Answer» Correct Answer - B | |
| 61. |
A copper bar of mass m sides under gravity on two smooth parallel rails l distance apart and set at angle `alpha` to the horizontal . At the lop , the rails are joined by a resistor R. Calculate the steady velocity of the bar the n when there is a unifrom magnetic field B perpedicular to the plane of the rails . |
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Answer» Correct Answer - `(mgR sin alpha)/(B^(2)l^(2))` |
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| 62. |
A rectangular loop of wire is placed perpendicualr to a uniform magnetic field and then spun around one of its sides at frequency f. The induced emf is a maximum when theA. Flux is zeroB. Flux is maximumC. Flux is half its maximum valueD. Derivative of the flux with respect to time is zero |
| Answer» Correct Answer - A | |
| 63. |
When a loop moves towards a stationary magnet with speed `v`, the induced emf in the loop is `E`. If the magnet also moves away from the lop with the same speed, then the emf inducted in the loop isA. `E`B. `2E`C. `E/2`D. zero |
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Answer» Correct Answer - D Relative velocity `=0` `:.` Charge in flux `=0` |
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| 64. |
A magnent moves towards a coil. Which of the following factors can affect the emf induced in the coil? |
| Answer» Correct Answer - D | |
| 65. |
A cylindrical bar magnet is kept along the axis of a circular coil. If the magnent is rotated about its axis, thenA. A current will be induced in the coilB. No current will be induced in the coilC. An emf and a current both will be induced in the coilD. None of these |
| Answer» Correct Answer - B | |
| 66. |
The induced emf produced when a magnet is inserted into a coil does not depend upon theA. Number of turns in the coilB. Resistance of coilC. Magnetic moment of the magnetD. Speed of approach of the magnet |
| Answer» Correct Answer - B | |
| 67. |
Which of the following is not an application of eddy currents?A. Electric power metersB. Induction furnaceC. LED lightsD. Magnetic brakes in trains |
| Answer» Correct Answer - C | |
| 68. |
Induction furnace is based on the heating effect ofA. self inductionB. mutual inductionC. eddy currentD. none of these |
| Answer» Correct Answer - C | |
| 69. |
A `40Omega` electric heater is connected to a `200V, 50Hz` main supply. The peak value of electric current flowing in the circuit is approx.A. 2.5AB. 5.0AC. 7AD. 10A |
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Answer» Correct Answer - D |
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| 70. |
A closed circuit consits of a source of constant and `E` and a choke coil of inductance `L` connected in series. The active resistance of the whole circuit is equal to `R`. At the moment `t = 0` the choke coil inductance was decreased abrupty `eta` times. FInd the current in the circuit as a function of time `t`.A. zeroB. E/RC. `(nE)/(R)`D. `(E)/(nR)` |
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Answer» Correct Answer - C Due to charge in inductance , magnetic flux remains constant . but ,`phi` = li= constant `L_(1)l_(1) = L_(2)l_(2)` `Here, L_(2) = (L_(1))/(n) and l_(1) = (E)/(R) therefore L_(2) = (L_(1)l_(1))/(L_(2)) = (nE)/(R)` |
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| 71. |
The current passing through a choke coil of inductance 4 Henry is decreasing at the rate of `3 amp//sec`. The e.m.f developed across the coil isA. `-8 V`B. `-10 V`C. `-12 V`D. `-6 V` |
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Answer» Correct Answer - C `e = -L(dI)/(dt) = 4(-3) = -12 V`. |
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| 72. |
A closed circuit consits of a source of constant and `E` and a choke coil of inductance `L` connected in series. The active resistance of the whole circuit is equal to `R`. At the moment `t = 0` the choke coil inductance was decreased abrupty `eta` times. FInd the current in the circuit as a function of time `t`. |
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Answer» Correct Answer - `(xi)/(R )[I-(eta-1)e^(etaR t)/(L)]` |
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| 73. |
When an alternating voltage of 100V, 50 Hz is applied to a choke coil, it takes a current of 10 A and the power is 500 W. What is the choke coil?A. `5sqrt(3) Omega`B. `4sqrt(3) Omega`C. `3 Omega`D. `3sqrt(3) Omega` |
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Answer» Correct Answer - A `Z = 100/10 = 10 Omega` `P = I^(2)R :. R = )/(I_2) = 500/100 = 5 Omega` `:. Z^(2) = R^(2) + X_(L)^(2)` `:. X_(L) = sqrt(Z^(2)-R^(2))= sqrt(100 - 25) = sqrt(75) = 5sqrt(3) Omega`. |
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| 74. |
Is there any device by which direct current can be controlled without any loss of energy ? Can a choke coil do so? |
| Answer» No, there is no device that can control d.c. without any energy loss. Even a choke coil cannot do so. | |
| 75. |
11 kW of electric power can be transmitted to a distant station at (i) 220 V (ii) 22000 V. Which of the two transmission modes be preferred and why ? Support your answer with calculations. |
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Answer» Here, `E_(p) xx I_(p) = 11 kW = 11000 W` (i) When power is transmitted at `E_(s) = 220 V` `I_(s) = (E_(p) I_(p))/( E_(s)) = (11000)/(220) = 50 A` Energy lost a heat `= I_(s)^(2) R = (50)^(2) R` `= 2500 R` watt (ii) When power is transmitted at `E_(s) = 22000 V` `I_(s) = (E_(p) I_(p))/(E_(s)) = (11000)/(22000) = 0.5 A` Energy lost as heat `= I_(s)^(2) = (0.5)^(2) R = 0.25 R` watt which is much less than energy loss in case (i). Therefore, power transmission should be done at 22000 V. |
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| 76. |
A sinusoidal waveform is given by `i=20 sin(6284 t+ 20^(@))A`. What is its period?A. 1 secondB. 1 nanosecondC. 1 microsecondD. 1 milisecond |
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Answer» Correct Answer - D `omega t = (2 pi t)/(T) = 6284 t` ` T = (2 xx 3.142)/(6284) = 1/1000 = 10^(-3) s`. |
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| 77. |
A resistance of `25 Omega` is connected to 100 V, 50 Hz a.c. source. What is the maximum instanteous current in the resistor?A. 6.66 AB. `2sqrt(2)A`C. `4sqrt(2)A`D. `4/(sqrt(2))A` |
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Answer» Correct Answer - C `I_(rms) = 100/25 = 4A` `:. I_(max)= sqrt(21_(rms)) = 4sqrt(2)A`. |
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| 78. |
The power factor of the circuit shown in figure isA. 0.4B. 0.8C. 0.9D. 0.7 |
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Answer» Correct Answer - B Power factor `=cosphi=(R)/(Z)` `=(R)/(sqrt(R^(2)+(X_(L)-X_(C))^(2)))` `=(80)/(sqrt(6400+3600))=(80)/(100)=0.8` |
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| 79. |
A circular coil and a bar magnet placed nearby are made to move in the same direction. The coil covers a distance of `1m` in `0.5 sec` and the magnet a distance of `2 m` in `1 sec`. The induced emf produced in the coilA. 0VB. 2VC. 1VD. 5V |
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Answer» Correct Answer - A `v_(b)=(d)/(t)=(2)/(1)=2m//s` `v_(c)=(d)/(t)=(1)/(0.5)=2m//s` Thus, both are moving with same velocity in same direction. Therefore emf production in the coil is zero. |
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| 80. |
Refer to above question , the potential difference between C and D is :A. 80 m VB. zeroC. 40 mVD. 60 mV |
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Answer» Correct Answer - c |
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| 81. |
A conducting ring of radius 2R rolls on a smooth horizontal conducting surface as shown in figure-5.295. A uniform horizontal magnetic field B is perpendicular to the plane of the ring. The potential of A with respect to O is: A. 2 BvRB. `(1)/(2)BvR`C. 8BvRD. 4BvR |
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Answer» Correct Answer - A |
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| 82. |
A thin semi-circular conducting ring of radius R is falling with its plane verticle in horizontal magnetic induction `(vec B)`. At the position MNQ the speed of the ring is v, and the potential difference developed across the ring is A. (a) zeroB. (b) `BVpiR^(2)//2` and `M` is at higher potentialC. ( c) `piRBV` and `Q` is at higher potentialD. (d) `2RBV` and `Q` is at higher potential |
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Answer» Correct Answer - D Induced emf: `B = ("effective length")v = B2Rv` |
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| 83. |
the uniform magnetic field perpendicular to the plane of a conducting ring of radius a change at the rate of `alpha`, thenA. All the points on the ring are at the same potentialB. The EMF induced in the ring is `pia^(2)alpha`C. Electric field intensity E at any point on the ring is zeroD. `E=(1)/(2)a alpha`. |
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Answer» Correct Answer - A::B::C |
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| 84. |
A thin semi-circular conducting ring of radius R is falling with its plane verticle in horizontal magnetic induction `(vec B)`. At the position MNQ the speed of the ring is v, and the potential difference developed across the ring is A. zeroB. `Bv pi R^(2)//2` and M is higher potencialC. `pi RBv ans Q is at higher potentialD. 2RBv ans Q is at higher potential |
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Answer» Correct Answer - D Induced emf produced across MNQ will be same as the induced emf produced in straight wire MQ. `:. E=Bvl=Bvxx2R` with Q at higher potential. |
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| 85. |
A semicircle conducting ring of radius R is placed in the xy plane, as shown in Fig. A uniform magnetic field is set up along the x-axis. No emf, will be induced in the ring if A. It moves along the x-axisB. It moves along the y-axisC. It moves along the z-axisD. It remains stationary |
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Answer» Correct Answer - A |
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| 86. |
The figure shows a conducting ring of radius `R`. A uniform steady magnetic field `B` lies perpendicular to the plane of the ring a circular region `r (lt R)`. If the resistance per unit length of the ring is `lamda`, then the current induced in the ring when its radius gets doubled is A. `(BR)/(lambda)`B. `(2BR)/(lambda)`C. Zero in 1, minimum in 2D. |
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Answer» Correct Answer - A |
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| 87. |
An athlete peddles a stationary tricycle whose pedals are atteached to a coil having 100 turns each of area `0.1 m^(2)`. The coil lying in XY plane is rotated in this plane at the rate of 50 rpm about the Y=axis in a region where a uniform magnetic field `vec B = (0.01) hat k` tesla is present. Find the (i) max. e.m.f. (ii) average e.m.f. generated in the coil over one complete rotation. |
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Answer» Here, `N = 100, A = 0.1 m^(2)`, `v = 50` rpm `= (5)/(6)` rps `vec B = (0.01) hat k T`, i.e., `B = 0.1 T` along z-axis. `e_(0) = NAB omega = 100 xx 0.1 xx 0.01 xx 2 pi xx (5)/(6)` `= 0.52 V` (ii) As the e.m.f. generated varies sinusoidally with time, so the average e.m.f. generated in the coil over one complete revolution is zero. |
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| 88. |
An electron moves along the line AB which lies in the same plane as a circular loop of conducting wire as shown in figure. What will be the direction of the current induced (if any) in the loop? A. The current will change the direction as the electorn passes byB. No current will be inducedC. The current will be clockwiseD. The current will be anticlockwise |
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Answer» Correct Answer - A An electron moving along the straight line is like a current carrying straight conductor, with a magnetic field around it. As a result, the flux linked with the coil will increase when the electron approaches the coil in going from left to right. But when it goes away, the flux will start decreasing. so the induced current will change the direction from anticlockwise as it passes by. |
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| 89. |
A metal coil of area `5 xx 10^(-3)m^(2)`, number of turns 100 and resistance `0.5 Omega` is lying horizontally at the bottom of a vessel made of an insultating material. A uniform magnetic field passing vertically through the coil changes from 0 to 0.8t in 0.2 s. What is induced current (in ampere) flowing through the coil?A. 2AB. 3AC. 4AD. 5A |
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Answer» Correct Answer - C The magnetic flux `phi = NAB` `:.` The induced e.m.f. = `e = -(d phi)/(dt)` `:. |e| = |(d phi)/(dt)| = NA (dB)/(dt)` But (`dB)/(dt) = 0.8/0.2 = 4 T//s` `:. e = NA(dB)/(dt)=100 xx 5 xx 10^(-3) xx 4 :. e = 2 V` `:.` induced current `I = e/R = 2/0.5 = 4 A`. |
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| 90. |
Magnetic flux linked with a stationary loop of resistance `R` varies with respect to time during the time period `T` as follows: `phi=aT(T-r)` Find the amount of heat generated in the loop during that time. Inductance of the coil is negligible.A. `(aT)/(3R)`B. `(a^(2)T^(2))/(3R)`C. `(a^(2)T^(2))`D. `(a^(2)T^92))/(3R)` |
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Answer» Correct Answer - D Given that `phi=at(T-t)` Induced e.m.f., `E=(dphi)/(dt)=(d)/(dt)[at(T-t)]` `=at(0-1)+a(T-t)` `=a(T-2t)` So, induced emf is also a function of time. :. Heat genrated in time `T` is `H=int_(0)^(T)(E^(2))/(R )dt=(a^(2))/(R )int_(0)^(T)(T-at)^(2)dt` `=(a^(2))/(R )int_(0)^(T)(E^(2))/(R )dt=(a^(2))/(R )int_(0)^(T)(T-at)^(2)dt` `=(a^(2))/(R )int_(0)^(T)(T^(2)+4t^(2)-4tT)dt=(a^(2)T(3))/(3R)` |
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| 91. |
A physicist works in a laboratory where the magnetic field is `2T`. She wears a necklace enclosing area `0.01 m^(2)` in such a way that the plane of the necklace is normal to the field and is having a resistance `R=0.01Omega`. Because of power failure, the field decays to `1T` in time `10^(-3)` seconds. The what is the total heat produced in her necklace?`(T=tesla)`A. 40 JB. 30 JC. 20 JD. 10 J |
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Answer» Correct Answer - D `t = 10^(-3) s, R = 10^(-2) Omega and A = 10^(-2) m^(2)` `e = (d phi)/(dt) = ((B_(1)-B_(2))A)/(dt)` `e = ((2-1)xx10^(-2))/(10^(-3)) = 10 V` `:.` The heat developed, `H = (V^2t)/R` `:. H = (10 xx 10 xx 10^(-3))/(10^(-2)) :. H = 10 J`. |
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| 92. |
A physicist works in a laboratory where the mahgnetic field is `2T`. She wears a necklace enclosing area `0.01 m^(2)` in such a way that the plane of the necklace is normal to the field and is having a resistance `R=0.01Omega`. Because of power failure, the field decays to `1T` in time `10^(-3)` seconds. The what is the total heat produced in her necklace?`(T=tesla)`A. `10J`B. `20J`C. `30J`D. `40J` |
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Answer» Correct Answer - A `H=(V^(2)t)/(R )` and `V=(N(B_(2)-B_(1))Acostheta)/(t)` `V=(1xx(1-2)xx0.01xxcos0^(@))/(10^(-3))=10V` |
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| 93. |
Figure show a square loop of side `0.5 m` and resistance `10Omega`. The magnetic field has a magnitude `B=1.0T`. The work done in pulling the loop out of the field slowly and uniformly in `2.0s` is A. (a) `3.125 xx 10^(-3)J`B. (b) `6.25 xx 10^(-4)J`C. ( c) `1.25 xx 10^(-2)J`D. (d) `5.0 xx 10^(-4)J` |
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Answer» Correct Answer - A (a) Speed of the loop should be `v = (l)/(t) = (0.5)/(2) = 0.25 m s^(-1)` Induced emf, `e = Bvl = (1.0)(0.25)(0.5) = 0.125 V` `:.` Current in the loop, `I = (e)/(R ) = (0.125)/(10)` `=1.25 xx 10^(-2)A` The magnetic force on the left arm due to the magnetic field is `F_(m) = ilB = (1.25 xx 10^(-2))(0.5)(1.0)` `=6.25 xx 10^(-3) N` To pull the loop uniformly an external force of `6.25 xx 10^(-3)` `N` towards right must be applied. `:.` `W = (6.25 xx 10^(-3) N)(0.5m) = 3.125 xx 10^(-3)J` |
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| 94. |
Figure show a square loop of side `0.5 m` and resistance `10Omega`. The magnetic field has a magnitude `B=1.0T`. The work done in pulling the loop out of the field slowly and uniformly in `2.0s` is A. `3.125xx10^(-3)J`B. `6.25xx10^(-4)J`C. `1.25xx10^(-2)J`D. `5.0x10^(-4)J` |
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Answer» Correct Answer - A Speed of the loop should be `v=(l)/(t)=(0.5)/(2)=0.25m//s` Induced emf, `eBvl=(1.0)(1.0)(0.25)(0.5)` `=0.125V` `:.` Current in the loop `i=(e)/(R )=(0.125)/(10)` `=1.25xx10^(-2)A` The magnetic force on the left arm due to the magnetic field is `F_(m)=ilB=(1.25xx10^(-2)(0.5)(1.0)` `=6.25xx10^(-3)N` to pull the loop uniform an external force of `6.25xx10^(-3)N` towards right must be applied. `:. W=(6.25xx10^(-3N)(0.5m)=3.125xx10^(-3)J` |
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| 95. |
Figures shows a square loop of side 1m and resistance `1Omega`. The magnetic field on left side of line PQ has a magnitude B=1.0T. The work done in pulling the loop out of the field uniformly in 1 s is A. 1JB. 10JC. 0.1JD. 100J |
| Answer» Correct Answer - A | |
| 96. |
A long straight wire carries a current `I_(0)`, at distance `a` and `b=3a` from it there are two other wires, parallel to the former one, which are interconnected by a resistance `R`(figure).A connector slides without friction along the wires with a constant velocity `v`.Assuming the resistance of the wires,the connector, the sliding contacts and the self-inductance of the frame to be negligible. The point of application (distance from the long wire) of magnetic force on sliding wire due to the long wire is `(2a)/(lnx)` from long wire.Then find out value of `x`. |
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Answer» Correct Answer - C `int`(Torque of elecentary force)=Torque of net force. `(mu_(0)I_(0))/(2pi)Iint(xdx)/x=F_(0)barx=(mu_(0)I_(0))/(2pi)I ln (b/a) barx rArr barx=(b-a)/(ln(b//a))rArrbarx=(2a)/(ln3) rArr x=3` |
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| 97. |
A square metallic wire loop of side `"0.1 m"` and resistance of `1Omega` is moved with a constant velocity in a magnetic field of `2wb//m^(2)` as shown in figure. The magnetic field field is perpendicular to the plane of the loop, loop is coonected to a network of resistances. what should be the velocity of loop so as to have a steady current of `1mA` in loop? A. (a) `2ms^(-1)`B. (b) `2ms^(-1)`C. ( c) `10ms^(-1)`D. (d) `20ms^(-1)` |
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Answer» Correct Answer - B Effective resistance is `4 Omega`. `I = (E)/(R ) = (Blv)/(R ) or v = (IR)/(Bl)` or `v = (1 xx 10^(-3) xx 4)/(2 xx 10 xx 10^(-2))ms^(-1)` `= 0.02 ms^(-1) = 2cm s^(-1)` |
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| 98. |
A square metal wire loop of side 10 cm and resistance 1 ohm is moved with a constant velocity `(v_0)` in a uniform magnetic field of induction `B=2 weber//m^(2)` as shown in the figure. The magnetic field lines are perpendicular to the plane to the loop (directed into the paper). The loop is connected to a network of resistors each of value 3 ohms. The resistances of hte lead wire OS and PQ are negligible. What should be the speed of the loop so as to have a steady current of 1 milliampere in the loop? Given the direction of current in the loop. A. 0.04 m/sB. 0.4 m/sC. 0.2 m/sD. 0.02 m/s |
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Answer» Correct Answer - D `I=(e)/(R)=(Blv)/(R)` `therefore v=(IR)/(Bl)=(1xx10^(-3)xx4)/(2xx0.1)=2xx10^(-2)m//s` |
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| 99. |
A square metallic loop of side `l` is placed near a fixed long wire carrying a current `i`(figure).The loop is moved towards right perpendicular to the wire with a speed `v` in the plane containing the wire and the loop.The `emf` induced in the loop when the rear end of the loop is at a distance `a=2l` from the wire is `(mu_(0)iv)/(xpi)`.Find out value of `x`. |
| Answer» Correct Answer - A::B | |
| 100. |
A square metal wire loop of side 10 cm and resistance 1 ohm is moved with a constant velocity `(v_0)` in a uniform magnetic field of induction `B=2 weber//m^(2)` as shown in the figure. The magnetic field lines are perpendicular to the plane to the loop (directed into the paper). The loop is connected to a network of resistors each of value 3 ohms. The resistances of hte lead wire OS and PQ are negligible. What should be the speed of the loop so as to have a steady current of 1 milliampere in the loop? Given the direction of current in the loop. A. `0.5cm//sec`B. `1cm//sec`C. `2cm//sec`D. `4cm//sec` |
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Answer» Correct Answer - C Induced emf `e=Bvl` `i=(Bvl)/(R_(eq))` `10^(-3)=(2xxvxx0.1)/(4) implies v=2cm//sec` |
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