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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.
| 1101. |
A conducting ring of mass `m = pi kg` and radius `R = (1)/(2)m` is kept on a flat horizontal surface (xy plane). A uniform magnetic field is switched on in the region which changes with time (t) as `vec(B) = (2hatj +t^(2) hatk)T`. Resistance of the ring is `r = pi Omega` and `g = 10 ms^(-2)`. (a) Calculate the induced electric field at the circumference of the ring at the instant it begins to topple. (b) Calculate the heat generated in the ring till the instant it starts to topple. |
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Answer» Correct Answer - (a) `10 V//m` (b) `(2pi)/(3)kJ` |
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| 1102. |
Two long co-axial solenoids have radii, and number of turns per unit length equal to `r_(1), r_(2)` and `n_(1), n_(2)` respectively where suffix 1 refers to the outer solenoid and 2 refers to the inner solenoid. Length of both is l. The current in the outer solenoid is made to grow as `I_(1) = kt` where t is time. Resistance of the wire used in inner solenoid is R. Write the current induced in the inner solenoid assuming that it is shorted. |
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Answer» Correct Answer - `i=(kpi mu_(0)n_(1)n_(2)r_(2)^(2)l)/(R) (1-e^(-t//tau))` where `tau = (pi mu_(0)n_(2)^(2) r_(2)^(2)l)/(R)` |
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| 1103. |
A rectangular conducting loop PQRS is kept in xy plane with its adjacent sides parallel to x and y axes. A mag- netic field `vec(B)` is switched on in the region which varies in posi- tion and time as `vec(B) = [B_(0) sin (ky – omega t)]hatk` . It was found that the emf induced in the loop is always zero. Express the length (L) of the loop in terms of constant k and `omega`. |
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Answer» Correct Answer - `L = n (2pi)/(k); n = 1,2,3,..` |
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| 1104. |
Which is the correct phasor diagram for an a.c. circuit containing only a pure capacitor?A. B. |
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Answer» Correct Answer - C The current leads the voltage by `90^(@)`. (c). |
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| 1105. |
A conducting rod of length L is carrying a constant current I. There exists a magnetic field B perpendicular to the rod. Due to the magnetic force the rod moved through a distance x in a direction perpendicular to the field (B) as well as its own length. The rod acquires a kinetic energy. A student says that a magnetic force ILB acted on the rod and it performed a work `W = ILB x` on the rod. But we know that magnetic force is always perpendicu- lar to the velocity of the charge and it cannot perform work on a moving charge. Which agency has actually spent energy to impart kinetic energy to the rod? |
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Answer» Correct Answer - The source that is driving current through the rod. |
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| 1106. |
In a rectangular circuit ABCD a capacitor having capacitance `C = 20 muF` is charged to a potential difference of 100 V. Resistance of circuit is `R = 10 Omega`. Another rectangular conducting loop PQRS is kept side by side to the first circuit with sides AB and PQ parallel and close to each other. The length and width of rectangle PQRS is a = 10 cm and b = 5 cm respectively and AB as well as BC is large compared to a. PQ is located near the centre of side AB with d = 5 cm. The loop PQRS has 25 turns and the wire used has resistance of `1 Omega m^(–1)`. The switch S is closed at time t = 0. Neglect self inductance of loops. (a) Find the current in ABCD at `t = 200 mus`. (b) Find the current in PQRS at `t = 200 mus`. |
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Answer» Correct Answer - (a) `3.7 A` (b) `35 mu A` |
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| 1107. |
A capacitor charged to 10 V is being discharged through a resistance R. At the end of 1 s, the voltage across the capacitor is `5 V`. What will be the voltage after 2 s? |
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Answer» The equation of discharging of capacitor is `q = q_(0) e^(-t//RC) , (q_(0))/(q) = e^(t//RC)` or `t = RC log_(e) ((q_(0))/(q)) = RC log_(e) ((V_(0))/(V))` At `t = 1 s, V = 5` volt from `V_(0) = 10 V` `:. 1 = RC log_(e) ((10)/(5)) = RC log_(e) 2` At `t = 2 s, V = ? V_(0) = 10 V` `:. 2 = RC log_(e) ((10)/(V))` Dividing (i) by (i), we get `2 = (log_(e) 10 - log_(e) V)/(log_(e) 2)` `2 log_(e) 2 = log_(e) 10 - log_(e) V` `log_(e) V = log_(e) 10 - log_(e) 2^(2) = log_(e) ((10)/(4))` `V = (10)/(4) = 2.5 V` |
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| 1108. |
The frequency of the output signal of an LC oscilliator circuit is 100 Hz, with a capacitance of `0.1 mu F`. If the value of the capacitor is increased to `0.2 mu F`, then the frequency of the output signal willA. be doubledB. be halfC. increase by `1/(sqrt2)`D. decrease by `1/(sqrt2)` |
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Answer» Correct Answer - D `f = 1/(2pisqrt(LC)) :. f prop 1/(sqrt(C)) as 1/(2pisqrt(L))` is consant. In this case `C_(1) = 0.1 mu F and C_(2) = o.2 mu F i.e. C_(2) = 2 C_(1)` `:. (f_2)/(f_1) = (sqrt(C_1))/(sqrt(C_2)) = 1/(sqrt(2))`. Thus `f_(2) = (f_1)/(sqrt(2))` Thus the frequency decreases by `1/(sqrt(2))`. |
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| 1109. |
In the circuit shown in Fig. the capacitor has capacitance `C = 20 muF` and is initially charged to `100 V` with the polarity shown. The resistor `R_(0)` has resistance `10 Omega`. At time `t = 0`, the switch is closed. The smaller circuit is not connected in any way to teh larger one. the wire of the smaller circuit has a resistance of `1.0 Omega m^(-1)` and contains `25` loops. the larger circuit is a rectangle `2.0 m` by `4.0 m`, while the smaller one has dimensions `a = 10.0 cm` and `b = 20.0 cm`. the distance `c`is `5.0 cm`. (The figure is not drawn to scale.) Both circuit are held stationary. Assume that only the wire nearest to teh smaller circuit produces. an appreciable magnetic field through it. The direction of current in the smaller circuit isA. (a) clockwiseB. (b) anticlockwiseC. ( c) always changes with timeD. (d) cannot be calculated |
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Answer» Correct Answer - B (b) The large circiut is a with a time constant of `tau = RC = (10 Omega)(20 xx 10^(-6)F) = 200 mus`. Thus, the currnt as a function of time is `I = ((100 V)/(10 Omega))e^((-t)/(200mus))` At `t = 200 ms`, we obtain `I = (10 A)(e^(-1)) = 3.7 A`. Assuming that only the long wire nearest the small loop produces an appreciable magnetic flux through the small loop, `Phi_(B) = int_( c)^(c + a)(mu_(0)ib)/(2pi)dr = (mu_(0)ib)/(2pi)1n(1 + (a)/(c ))` ltbr. So the emf induced in the small loop at `t = 200 ms` `epsilon = -(dPhi)/(dt) = -(mu_(0)ib)/(2pi)1n(1 + (a)/(c ))(di)/(dt)` `= -(((4pi xx 10^(-7)(Wb))/(A xx m^(2)))(0.200 m))/(2pi) xx 1n (3.0)(-(3.7 A)/(200 xx 10^(-6) s))` Thus, the induced current in the small loop is `i = (epsilon)/(R ) = (0.81 mV)/(25(0.600 m)(1.0 Omega//m)) = 54 muA`. ltbr. Initially current in large loop is maximum and afterwards decreases. Hence flux through the smaller loop decreases with time. The induced current will act to oppose the decreases in flux from the large loop. Thus, the induced current flows counterclockwise . |
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| 1110. |
In the circuit shown in Fig. the capacitor has capacitance `C = 20 muF` and is initially charged to `100 V` with the polarity shown. The resistor `R_(0)` has resistance `10 Omega`. At time `t = 0`, the switch is closed. The smaller circuit is not connected in any way to teh larger one. the wire of the smaller circuit has a resistance of `1.0 Omega m^(-1)` and contains `25` loops. the larger circuit is a rectangle `2.0 m` by `4.0 m`, while the smaller one has dimensions `a = 10.0 cm` and `b = 20.0 cm`. the distance `c`is `5.0 cm`. (The figure is not drawn to scale.) Both circuit are held stationary. Assume that only the wire nearest to teh smaller circuit produces. an appreciable magnetic field through it. The curent in the smaller circuit `200 mus` after closing `S` isA. (a) `54 muA`B. (b) `10 muA`C. ( c) `15 muA`D. (d) `36 muA` |
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Answer» Correct Answer - A (a) The large circiut is a with a time constant of `tau = RC = (10 Omega)(20 xx 10^(-6)F) = 200 mus`. Thus, the currnt as a function of time is `I = ((100 V)/(10 Omega))e^((-t)/(200mus))` At `t = 200 ms`, we obtain `I = (10 A)(e^(-1)) = 3.7 A`. Assuming that only the long wire nearest the small loop produces an appreciable magnetic flux through the small loop, `Phi_(B) = int_( c)^(c + a)(mu_(0)ib)/(2pi)dr = (mu_(0)ib)/(2pi)1n(1 + (a)/(c ))` ltbr. So the emf induced in the small loop at `t = 200 ms` `epsilon = -(dPhi)/(dt) = -(mu_(0)ib)/(2pi)1n(1 + (a)/(c ))(di)/(dt)` `= -(((4pi xx 10^(-7)(Wb))/(A xx m^(2)))(0.200 m))/(2pi) xx 1n (3.0)(-(3.7 A)/(200 xx 10^(-6) s))` Thus, the induced current in the small loop is `i = (epsilon)/(R ) = (0.81 mV)/(25(0.600 m)(1.0 Omega//m)) = 54 muA`. ltbr. Initially current in large loop is maximum and afterwards decreases. Hence flux through the smaller loop decreases with time. The induced current will act to oppose the decreases in flux from the large loop. Thus, the induced current flows counterclockwise . |
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| 1111. |
A capacitor of capacitance C is having a charge `Q_(0)`. It is connected to a pure inductor of inductance L. The inductor is a solenoid having N turns. Find the magnitude of magnetic flux through each of the N turns in the coil at the instant charge on the capacitor becomes `(Q_(0))/(2)`. |
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Answer» Correct Answer - `(1)/(2) sqrt((3L)/(C)) (Q_(0))/(N)` |
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| 1112. |
In the circuit shown in Fig. the capacitor has capacitance `C = 20 muF` and is initially charged to `100 V` with the polarity shown. The resistor `R_(0)` has resistance `10 Omega`. At time `t = 0`, the switch is closed. The smaller circuit is not connected in any way to teh larger one. the wire of the smaller circuit has a resistance of `1.0 Omega m^(-1)` and contains `25` loops. the larger circuit is a rectangle `2.0 m` by `4.0 m`, while the smaller one has dimensions `a = 10.0 cm` and `b = 20.0 cm`. the distance `c`is `5.0 cm`. (The figure is not drawn to scale.) Both circuit are held stationary. Assume that only the wire nearest to teh smaller circuit produces. an appreciable magnetic field through it. The current in the circuit `200 ms` after closing `S` isA. (a) `(5)/(e) A`B. (b) `(2)/(e)A`C. ( c) `(15)/(e)A`D. (d) `(10)/(e)A` |
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Answer» Correct Answer - D (d) The large circiut is a with a time constant of `tau = RC = (10 Omega)(20 xx 10^(-6)F) = 200 mus`. Thus, the currnt as a function of time is `I = ((100 V)/(10 Omega))e^((-t)/(200mus))` At `t = 200 ms`, we obtain `I = (10 A)(e^(-1)) = 3.7 A`. Assuming that only the long wire nearest the small loop produces an appreciable magnetic flux through the small loop, `Phi_(B) = int_( c)^(c + a)(mu_(0)ib)/(2pi)dr = (mu_(0)ib)/(2pi)1n(1 + (a)/(c ))` ltbr. So the emf induced in the small loop at `t = 200 ms` `epsilon = -(dPhi)/(dt) = -(mu_(0)ib)/(2pi)1n(1 + (a)/(c ))(di)/(dt)` `= -(((4pi xx 10^(-7)(Wb))/(A xx m^(2)))(0.200 m))/(2pi) xx 1n (3.0)(-(3.7 A)/(200 xx 10^(-6) s))` Thus, the induced current in the small loop is `i = (epsilon)/(R ) = (0.81 mV)/(25(0.600 m)(1.0 Omega//m)) = 54 muA`. ltbr. Initially current in large loop is maximum and afterwards decreases. Hence flux through the smaller loop decreases with time. The induced current will act to oppose the decreases in flux from the large loop. Thus, the induced current flows counterclockwise . |
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| 1113. |
A capacitor of capacitance `0.5 mu F` is discharged through a resistance. Find the value of resistance if half the charge on capacitor escapes in one minute. |
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Answer» Hene, `C 0.5 mu F = 0.5 xx 10^(-6) F, R = ?` `q = (q_(0))/(2), t = 1 min. = 60 sec`. As `q = q_(0) e^(-t//RC) :. (q_(0))/(2) = q_(0) e^(-60 // RC)` `log_(e) 1 - log_(e) 2 = - (60)/(RC) log_(e) e` `0 - 2.303 log_(10) 2 = - (60)/(RC) xx 1` `2.303 xx 0.3010 = (60)/(R(0.5 xx 10^(-6)))` `:. R = (60)/(0.5 xx 10^(-6) xx 2.303 xx 0.3010)` ` = 1.73 xx 10^(8) ohm` |
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| 1114. |
A ring is made of a nearly superconducting mate- rial. Inductance of the ring is `L = 0.5 H`. A current allowed to decay in the ring was observed to remain constant for a month. The instrument used to measure current could detect any change if it is greater than `1%`. Estimate the resistance of the ring considering it as a L R circuit with decaying current. |
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Answer» Correct Answer - `1.9 xx 10^(-9) Omega` |
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| 1115. |
In the circuit shown, switch `S_(2)` is open and `S_(1)` is closed since long. Take `E = 20 V, L = 0.5 H` and `R = 10 Omega`. Find the rate of change of energy stored in the magnetic field inside the inductor, immediately after `S_(2)` is closed. |
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Answer» Correct Answer - `-40 J//s` |
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| 1116. |
In the circuit shown, the switch ‘S’ has been closed for a long time and then opens at t = 0. (a) Find the current through the inductor just before the switch is opened. (b) Find the current I a long time after the switch is opened. (c) Find current I as a function of time after the switch is opened. Also write the current through the cell as a function of time. |
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Answer» Correct Answer - (a) 6A (b) 44 (c) `I = 4 + 2e^(-5t)`, current through cell `= (21)/(4) + 2e^(-5t)` |
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| 1117. |
A coil having inductance 2.0 H and resistance `20 Omega` is connected to a battery of emf 4.0 V. Find (a) the current a the instant 0.20 s after the connection Is made and (b) the magnetic energy at this instant. |
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Answer» Correct Answer - A::B::C (a)`I=I_(0)(1-e^(-t//tau))` where `i_(0)=4/20=1/5` `tau=L/R=2/20=1/10` and `t=0.2 sec` (b)`E=1/2L I^(2)` |
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| 1118. |
A solenoid of resistance `50 Omega` and inductance 80 H is connected to a 200 V battery, How long will it take for the current to reach 50% of its final equlibrium value ? Calculate the maximum enargy stored ? |
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Answer» Here, `R = 50 Omega, L = 80 H`, `E = 200 V, t = ?` `I = (50)/(100) I_(0)` `tau = (L)/( R) = (80)/(50) = (8)/(5)s , I_(0) = (E)/(R ) = (200)/(50) = 4 A` From `I = I_(0) (1 - e^(-t//tau))`, `t = tau log_(e) 2 = (8)/(5) xx 0.693 = 1.109 s` Maximum energy stored `= (1)/(2) LI_(0)^(2) = (1)/(2) xx 80 xx (4)^(2) = 640 J` |
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| 1119. |
In the figure shown as uniform magnetic field `|B|=0.5T` is perpendicular to the plane of circuit. The sliding rod of length `l=0.25 m` moves uniformly with constant speed `v=4ms^-1`. If the resistance of the sllides is `2Omega`, then the current flowing through the sliding rod is A. `0.1A`B. `0.17A`C. `0.08`D. `0.03A` |
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Answer» Correct Answer - A `e=Bvl=0.5xx4xx0.25=0.5V` `12Omega` and `4Omega` are parallel.Hence, their net resistance `R=3Omega` `i=e/(R+r)=0.5/(32)=0.1A` |
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| 1120. |
A coil is placed in a magnetic field directed downward and increasing from 0 to 18 T in 0.1 s. area of coil is 2 `m^(2)` and resistance 5 `Omega`. Induced current will beA. 72 A anticlockwise directionB. 27 A anticlockwise directionC. 72 A clockwise directionD. 27 A clockwise direction |
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Answer» Correct Answer - A `I=(e)/(R)=(1)/(R)(dphi)/(dt)` `=(A)/(R)(dB)/(dt)=72A` Current induced will oppose the increase of downward magnetic field, so current will be anticlockwise. |
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| 1121. |
A large coil of 10 turns has a resistance 2 `Omega`. It is kept in a magnetic field of 0.5 T. if the coil is pulled out of the magnetic field uniformly such that its area coming out of the magnetic field is 200 `cm^(2)//s`, the current induced in it isA. 5 mAB. 50 mAC. 2.5 mAD. 25 mA |
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Answer» Correct Answer - B `I=(e)/(R)` `=(n)/(R)(dphi)/(dt)=(nB)/(R)(dA)/(dt)=50mA`. |
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| 1122. |
A circular coil having 20 turns, each of radius 8 cm, is rotating about its vertical diameter with an angular speed of 50 `rad//s` in a uniform horizontal magnetic field of 30 mT. Obtain the maximum, average and rms value of e.m.f. induced in the coil. If the coil forms a closed loop of resistance 10 ohm, how much power is dissipated as heat |
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Answer» Here, `N = 20, r = 8 cm = 8 xx 10^(-2) m`, `omega = 50 rad s^(-1)` `B = 30 mT = 30 xx 10^(-3) T, E_(0) = ?` `E_(av) = ?, E_(v) = ?` `R = 10 Omega, P = ?` `E_(0) = N AB omega = (N pi r^(2)) B omega` `= 20 xx 3.14 xx (8 xx 10^(-2))^(2) xx 30 xx 10^(-3) xx 50` `= 0.60 V` Average value of alt. emf over full cycle, `E_(av) = 0` `E_(v) = (E_(0))/(sqrt2) = (0.60)/(1.414) = 0.42 V` Power dissipated, `P = (E_(v)^(2))/(R ) = ((0.42)^(2))/(10) = 1.76 xx 10^(-2) W` |
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| 1123. |
The distance between the edges of the wings of an aeroplane is 30 metre. It is landing down with a velocity of 300 `km//hr`. If while landing, the wings of the aeroplane be east-west, find out the potentail difference between the edges of the wings. What will happen if the wings are along north south. Take `H = 0.4` gauss. |
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Answer» When wings are along east-west, they intercept H. `:. E = B l upsilon = H l upsilon` `= (0.4 xx 10^(-4)) xx 30 xx (3 xx 10^(5))/(60 xx 60) = 0.1 V` However, when wings are along north-south, H. Intercept neither H nor V.So, in that case, induced e.m.f. = 0 |
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| 1124. |
A circular disc of radius 20 cm is rotating with a constant angular speed of `2.0 rad//s` in a uniform magnetic field of 0.2 T. Find the e.m.f. induced between the centre and rim of the disc. Given magnetic field is along the axis of rotation of disc. |
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Answer» Here, `r = 20 cm = (1)/(5) m` `omega = 2.0 rad//s, B = 0.2 T, e = ?` `e = (1)/(2) B omega r^(2) = (1)/(2) xx 0.2 xx 0.2 xx ((1)/(5))^(2)` `= 8 xx 10^(-3) V = 8 mV` |
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| 1125. |
A coil of metal wire is kept stationary in a non-uniform magnetic field. An e.m.f. Is induced in the coil.A. an e.m.f. and current are both induced in the coilB. a current but no e.m.f. is induced in the coilC. an e.m.f. but no current is induced in the coilD. neither e.m.f nor current is induced in the coil |
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Answer» Correct Answer - D Since the coil is kept, stationary in a non uniform magnetic field, there is no change in the flux. Hence no e.m.f. or current will be induced in the coil. |
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| 1126. |
A coil of metal wire is kept stationary in a non-uniform magnetic field. An e.m.f. Is induced in the coil. |
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Answer» For indeced emf to develop in a coil, the magnetic flux through it must change. But in the number of magnetic lines of force through the coil is not changing. Therefore, the statement is false. |
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| 1127. |
A square loop of side `12 cm` with its sides parallel to `x` and `y`- axes is moved with a velocity `8 cm//s` along positive `x`-direction in an environment containing magnetic field along `+ve` z-direction. The field has a gradient of `10^(-3) "tesla"//"em"` along `-ve` x-direction (increasing along `-ve` x-axis) and also decreases with time at the rate of `10^(-3) "tesla"//s`. The emf induced in the loop is |
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Answer» Here, area of loop, `A = (12 xx 10^(-2))^(2) = 144 xx 10^(-4) m^(2)` , velocity, `upsilon = 8 cm s^(-1) = 8 xx 10^(-2) ms^(-1)` `(db)/(dx) = 10^(-3) T cm^(-1)` `(dB)/(dt) = 10^(-3) T s^(-1), R = 4.5 xx 10^(-3) ohm` Induced e.m.f. e = ? Rate of change of magnetic flux due to explicit time variation in `B = A ((dB)/(dt))` `= (144 xx 10^(-4)) xx 10^(-3) Wb s^(1) = 1.44 xx 10^(-5) Wb s^(-1)`. Rate of change of magnetic flux due to motion of the loop in non-uniform magnetic field `= A ((dB)/(dx)) ((dx)/(dt))` `= (144 xx 10^(-4)) xx 10^(-3) xx 8 = 11.52 xx 10^(-5) Wb//s` As both the effects cause a decrese in magnetic flux along the positive Z-direction, therefore, they add up. `:.` Total induced e.m.f. `e = 1.44 xx 10^(-5) + 11.52 xx 10^(-5) = 12.96 xx 10^(-5) V` Induced current, `= (e)/(R ) = (13.96 xx 10^(-5))/(4.5 xx 10^(-3)) = 2.88 xx 10^(-2)` The direction of induced current is such as to increase the flux through the loop along positive Z-direction. For example, for the observer, if the loop moves of the right, the current will be seen to be anticlockwise. |
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| 1128. |
An e.m.f. Can be induced between the two ends of a straight copper wire when it is moved through a uniform magnetic field. |
| Answer» True: A copper wire is at rest, the average velocity of each electron is zero. But when the wire is in motion, the electons have a net velocity in the direction of motion. | |
| 1129. |
A pair of adjacent coils has a mutual inductance of 2.5 H. If the current in one coil changes from 0 of 40 A in 8.0 s, then the change in flux linked with the other coil is.A. 100 WbB. 120 WbC. 200 WbD. 250 Wb |
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Answer» Correct Answer - A Given, `M=2.5H,(dI)/(dt)=(40-0)/(0.8)=50 As^(-1)` Also, `epsi-M=(dI)/(dt)=(dphi)/(dt)` or `d phi=MdI=2.5(40-0)=100 Wb` |
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| 1130. |
An electric potential difference will be induced between the ends of the conductor shown in the diagram, when the conductor moves in the direction A. `P`B. `Q`C. `L`D. `M` |
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Answer» Correct Answer - D cionductor cuts the flux only when, if it moves in the direction of `M`. |
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| 1131. |
A vertical matallic pole falls down through the plane of magnetic meridian. Will any e.m.f. be induced between its ends ? |
| Answer» No, because the pole intercepts neither H nor V. | |
| 1132. |
An electric potential difference will be induced between the ends of the conductor shown in the figure, if the conductor moves in the direction shown by A. `P`B. `Q`C. `L`D. `M` |
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Answer» Correct Answer - D Conductor cuts the flux only when, if it moves in the direction of `M`. |
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| 1133. |
A horizontal wire `0.8 m` long is falling at a speed of `5 m//s` perpendicular to a uniform magnetic field of `1.1 T`, which is directed from east to west. Calculate the magnitude of the induced ernf. Is the north or south end of the wire positive?A. `0.15V`B. `1.5mV`C. `1.5 V`D. `15.0V` |
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Answer» Correct Answer - B Induced e.m.h.`=B//v=0.3xx10^(-4)xx10xx5` `=1.5xx10^(-3)V=1.5mV` |
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| 1134. |
A conductor of `3 m` in length is moving perpendicularly to magnetic field of `10^(-4)` tesla with the speed of `10^(2)m//s`, then the e.m.f. produced across the ends of conductor will beA. `0.03 volt`B. `0.3volt`C. `3xx10^(-3) volt`D. `"3 volts"` |
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Answer» Correct Answer - B `e=Bvl=3xx10^(-3)xx10^(2)xx=0.3` volt |
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| 1135. |
A conductor of `3 m` in length is moving perpendicularly to magnetic field of `10^(-4)` tesla with the speed of `10^(2)m//s`, then the e.m.f. produced across the ends of conductor will beA. `7.25 V`B. `3.75 V`C. `1.25 V`D. `2.52 V` |
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Answer» Correct Answer - D `e=Bvlimpliese=0.9xx7xx0.4=2.52 V` |
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| 1136. |
Coefficient of coupling between two coils of self-inductances `L_(1)` and `L_(2)` is unity. It meansA. `50%` flux of `L_(1)` is linked with `L_(2)`B. `100%` flux of `L_(1)` is linked with `L_(2)`C. `sqrt(L_(1))` time of flux of `L_(1)` is linked with `L_(2)`D. None of the above |
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Answer» Correct Answer - B Two coils ace said to be magnetically coupled, if full or a part of the flux produced by one link with the other. Let `L_(1)` and `L_(2)` be the self-inductances of the coils and M be their mutual inductances, then `k=(M)/(sqrt(L_(1)L_(2))` When `100%` flux produced by one coil links with the other, then mutual inductance between the two is maximum and is given by `Msqrt(L_(1)L_(2))` |
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| 1137. |
A square loop `ABCD` of edge a moves to the right with a velocity `v` parallel to `AB` . There is a uniform magnetic field of magnitude `B` , direction into the paper, in the region between `PQ` and `RS` only. I, II and III are three ppositions of the loop. (i) The emf induced in the loop has magnitude `B` a `v` in all three position (iii) Induced emf is anticlockwise in position II (iv) The induced emf is clockwise in position III A. `(i), (iii)`B. `(ii), (iii), (iv)`C. `(i), (ii)`D. `(iii), (iv)` |
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Answer» Correct Answer - B In position II, change in flux=0, emf=0 In position I and III, `e=Bav` In position I, flux is incrreasing, hence anticlockwise emf In position III, flux is decreasing, hence clockwise emf |
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| 1138. |
A square loop `ABCD` of edge a moves to the right with a velocity `v` parallel to `AB` . There is a uniform magnetic field of magnitude `B` , direction into the paper, in the region between `PQ` and `RS` only. I, II and III are three ppositions of the loop. (i) The emf induced in the loop has magnitude `B` a `v` in all three position (iii) Induced emf is anticlockwise in position II (iv) The induced emf is clockwise in position III A. (i), (iii)B. (ii), (iii), (iv)C. (i), (ii)D. (iii),(iv) |
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Answer» Correct Answer - B In position II, change in flux = 0, emf = 0. In position I, and III, e = Bav In position I, flux is increasing, hence anti-clockwise emf. If position III, flux is decreasing, hence clockwise emf. |
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| 1139. |
A rectangular, a square , a circular and an elliptical loop, all in the `(x-y)` plane, are moving out of a uniform magnetic field with a constant velocity `vec(v)=vhati` . The magnetic field is directed along the negative `z`-axis direction. The induced emf, during the passage of these loops , out of the field region, will not remain constant for :A. The rectangular, circular and elliptical loopB. the circular and the elliptical loopsC. only the elliptical loopD. any of the four loop |
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Answer» Correct Answer - B Area coming out per second from the magnetic field is not constant for elliptical and circular loops, so induced emf, during the passage of these loops, out of the field region will not remain constant for the circular and the elliptical loops. |
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| 1140. |
A conductor of `3 m` in length is moving perpendicularly to magnetic field of `10^(-4)` tesla with the speed of `10^(2)m//s`, then the e.m.f. produced across the ends of conductor will beA. `0.5 volt`B. `0.1 volt`C. `1 volt`D. `2 volt` |
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Answer» Correct Answer - C `e=Bvl=0.5xx2xx1=1V` |
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| 1141. |
The current in the primary coil at time t is given by `I=(8t^(2)-4)` Ampere. If the e.m.f induced in the secondary coil is given by `e_(s)=32 xx 10^(-3)` t volt, then the mutual inductance between the two coil isA. 1 millihenryB. 2mHC. 5 mHD. 10 mH |
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Answer» Correct Answer - B `I = 8t^(2)-4 :. (dI)/(dt) = 16 t :. E = M(dI)/(dt)` `:. 32 xx 10^(-3) t = M xx 16 t :. M = 2 xx 10^(-3) H`. |
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| 1142. |
A rectangular, a square , a circular and an elliptical loop, all in the `(x-y)` plane, are moving out of a uniform magnetic field with a constant velocity `vec(v)=vhati` . The magnetic field is directed along the negative `z`-axis direction. The induced emf, during the passage of these loops , out of the field region, will not remain constant for :A. the rectangular, circular and elliptical loopsB. the circular and the elliptical loopsC. only the elliptical loopD. any of the four loops |
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Answer» Correct Answer - B Induced emf, `|e|=(dphi)/(dt)=(BdS)/(dt)` Now, as the square loop and rectangular loop move out of magnetic field, `(BdS)/(dt)` is constant, therefore `|e|` is constant. But in case of circular and elliptical loops, `(dS)/(dt)` changes. Therefore, `|e|` does not remain constant. |
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| 1143. |
A long horizontal metallic rod with length along the east-west direction is falling under gravity. The potential difference between its two ends willA. be zeroB. be constantC. increase with timeD. decrease with time |
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Answer» Correct Answer - C Induced emf, e = Bvl `therefore" "e prop v` Also, v = 0 + gt `therefore" "e prop "gt"` So, potential difference between its two ends will increase with time. |
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| 1144. |
A coil is suspended in a uniform magnetic field, with the plane of the coil parallel to the magnetic lines of force. When a current is passed through the coil it starts oscillating, It is very difficult to stop. But if an aluminium plate is placed near to the coil, it stops. This is due to :A. development of air current when the plate is placed.B. induction of electrical charge on the plateC. shielding of magnetic lines of force as aluminium is a paramagnetic material.D. Electomagnetic induction in the aluminium plate giving rise to electromagnetic damping. |
| Answer» Correct Answer - D | |
| 1145. |
A coil is suspended in a uniform magnetic field, with the plane of the coil parallel to the magnetic lines of force. When a current is passed through the coil it starts oscillating, It is very difficult to stop. But if an aluminium plate is placed near to the coil, it stops. This is due to :A. Electronagnetic induction in the aluminium plate giving rise to electroagnetic damping by eddy currentsB. Development of air current when the plate is placedC. Induction of electrical charge on the plateD. Shiclding of magnetic lines of force as aluminium is a paramagnetic material. |
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Answer» Correct Answer - B |
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| 1146. |
A coil is suspended in a uniform magnetic field, with the plane of the coil parallel to the magnetic lines of force. When a current is passed through the coil it starts oscillating, It is very difficult to stop. But if an aluminium plate is placed near to the coil, it stops. This is due to :A. induction of electrical charge on the plateB. shielding of magnetic lines of force as aluminium is a paramagnetic materialC. electromagnetic induction in the aluminium plate giving rise to electromagnetic dampingD. development of air current when the plate is placed |
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Answer» Correct Answer - C Electromagnetic induction in the aluminium plate gives rise to electromagnetic damping and hance the coil stops oscillating. |
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| 1147. |
Pure inductance of `3.0 H` is connected as shown below. The equivalent inductance of the circuit is A. 1 HB. 2 HC. 3 HD. 9 H |
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Answer» Correct Answer - A In the given circuit, three inductances are in parallel, their equivalent inductance is given by `(1)/(L_(eq))=(1)/(3)+(1)/(3)+(1)/(3)or L_(eq)=(3)/(3)=1H` |
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| 1148. |
The equivalent inductance of two inductances is `2.4` henry when connected in parallel and `10` henry when connected in series. The difference between the two inductance isA. `2` henryB. `3` henryC. `4` henryD. `5` henry |
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Answer» Correct Answer - A `L_(s)=L_(1)+L_(2)=10H` …..(i) `L_(p)=(L_(1)L_(2))/(L_(1)+L_(2))=2.4H` ……(ii) On solving (i) and(ii)`L_(1)L_(2)=24` Also `(L_(1)-L_(2))^(2)=(L_(1)+L_(2))^(2)-4L_(1)L_(2)` `implies(L_(1)-L_(2))^(2)=(10)^(2)-4xx24=4impliesL_(1)-L_(2)=2H` |
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| 1149. |
A circular copper disc of 10 cm radius rotates at `20 pi` `radian//sec` about an axis through its centre and perdendicular to the disc. A uniform magnetic field of 0.2 T acts perpendicular to the disc. Calculate the potential difference developed between axis of the disc and the rim. What is the induced current if resistance of disc is 2 ohm ? |
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Answer» Here, `r = 10 cm = 10^(-1) m, omega = 20 pi rad//s` `B = 0.2 T, e = ?` Potential difference develop between the axis of the disc and its rim, `e = (1)/(2) B r^(2) omega` `e = (1)/(2) xx 0.2 (10^(-1))^(2) xx 2 pi = 0.0629 V` Induced current, `i = (e)/(R ) = (0.0628)/(2) = 0.0314 A` |
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| 1150. |
The equivalent inductance of two inductances is `2.4` henry when connected in parallel and `10` henry when connected in series. The difference between the two inductance isA. 8 H, 2HB. 6H, 4HC. 5H, 5HD. 7H, 3H |
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Answer» Correct Answer - B In series connection `L_(1)+L_(2)=10 H " ".....(ii)` and in parallel connection `(L_(1)L_(2))/((L_(1)+L_(2)))=2.4H" "...(ii)` Subsitituting the vlaue of `(L_(1)+L_(2))` form (i) into, (ii), we get `L_(1)L_(2)=(2.4)(L_(1)+L_(2))=2.4xx10=24` `(L_(1)L_(2))=(L_(1)+L_(2))^(2)-4L_(1)L_(2)` `L_(1)-L_(2)=[(10)^(2)-4xx24]^(1//2)=2H" "...(iii)` Solvin (i) and (iii), we get `L_(1)=6H,L_(2)=4H` |
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