InterviewSolution
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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.
| 1. |
Two circular, similar, coaxial loops carry equal currents in the same direction. If the loops are brought nearer, what will happen?A. Current will increase in each loopB. Current will decrease in each loopC. Current will remain same in each loopD. Current will increase in one and decrease in the other |
| Answer» Correct Answer - B | |
| 2. |
A conducting rod of length `2l` is rotating with constant angular speed `w` about its perpendicular bisector. A uniform magnetic field `B` exists parallel to the axis of rotation. The e.m.f. induced between two ends of the rod is A. `B omega l`B. `(1)/(2)B omega l^(2)`C. `(1)/(8)B omega l^(2)`D. Zero |
| Answer» Correct Answer - D | |
| 3. |
An inductor of `2` henry and a resistance of `10` ohms are connected in series with a battery of `5` volts. The initial rate of change of current isA. `0.5 amp//sec`B. `2.0 amp//sec`C. `2.5 amp//sec`D. `0.25 amp//sec` |
| Answer» Correct Answer - C | |
| 4. |
A thin wire of length `L` is connected to two adjacent fixed points and carries a current `I` in the clockwise direction , as shown in the figure. When the system is put in a uniform magnetic field of strength `B` going into the plane of the paper , the wire takes the shape of a circle . The tension in the wire is A. `IBL`B. `(IBL)/(pi)`C. `(IBL)/(2pi)`D. `(IBL)/(4pi)` |
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Answer» Correct Answer - C `L = 2pi R, :. R = (L)/(2pi), 2T sin (d theta) = d theta` For small angles, `sin (d theta) = d theta` `:. 2T(d theta) = I(dL)B sin 90^(@), = I (2R. d theta).B` `:. T = IRB = (ILB)/(2pi), :.` Correct option is (c ) |
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| 5. |
A short-circuited coil is placed in a time-varying magnetic field. Electrical power is dissipated due to the current induced in the coil. If the number of turns were to be quadrupled and the wire radius halved, the electrical power dissipated would be |
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Answer» Current is induced in the short - circuited coil due to the imposed time-varying magnetic field. Power `P = (e^(2))/(R )` , Here `e = -(d phi)/(dt)` where `phi = NBA` and `R = (rho l)/(pi r^(2))` where `l` and `r` length and radius of the wire. `:. P = (pi r^(2))/(rho l)[(d)/(dt)NBA]^(2)` or `P = (pi r^(2))/(rho l) N^(2)A^(2)((dB)/(dt))^(2)` or `P =("constant") (N^(2)r^(2))/(l)` when `r_(2) = (r_(1))/(2)` then `t_(2) = 4l_(1)` `:. (P_(2))/(P_(1)) = ((4N)^(2))/(N^(2)) xx ((r)/(2r))^(2) xx ((l)/(4l))` `:. (P_(2))/(P_(1)) = (16 N^(2) xx r^(2) xx l)/(N^(2) xx 4r^(2) xx 4l)` or `(P_(2))/(P_(1)) = (1)/(1)` `:.` Power dissipated is the same. |
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| 6. |
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. `10J`B. `20J`C. `30J`D. `40J` |
| Answer» Correct Answer - A | |
| 7. |
a physicist works in a laboratory where the magnetic field is `2T`. She wears a necklace enclosing of an area `100 cm^(2)` of field and having a resistance of `0.1 Omega`. Because of power failure, the field decays to `1 T` in millisecond. The electric charge circulated in the necklace assuming that the magnetic field is perpendicular to area covered by the necklace isA. `0.01 C`B. `0.001 C`C. `0.1 C`D. `1.0 C` |
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Answer» Correct Answer - C `q = (N(B_(1)-B_(2))A)/(R )=(1(2-1)10^(-2))/(0.1)=0.1C` |
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| 8. |
The linear loop has an area of `5 xx 10^(-4)m^(2)` and a resistance oof `2 Omega`. The larger circular loop is fixed and has a radius of `0.1m`. Both the loops are concentric and coplanner. The smaller loop is rotated with an angular velocity `omegarads^(-1)` about its dismeter. The magnetic flux with the smaller loop is A. `2pi xx 10^(-6) "weber"`B. `pi xx 10^(-9) "weber"`C. `pi xx 10^(-9) cos omega t "weber"`D. zero |
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Answer» Correct Answer - C Magnetic field due to larger loop `=(mu_(0)1)/(2r)=(4pixx10^(-7)xx1)/(2xx0.1)T=2pixx10^(-6)T` Now, magnetic flux linked withthe smaller loop, `phi` `=NBA cos omega t=1xx2pixx10^(-6) xx5xx10^(-4) cos omegat` |
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| 9. |
A non-conducting disk of radius `R` is rotating about its own axis with constant angular velocity `omega` in a perpendicular uniform magnetic field `B` as shown in figure. The emf induced between centre and rim of disk is A. `pi omega B R^(2)`B. `omega B R^(2)`C. `pi omega R//2`D. `omega B R^(2)//2` |
| Answer» Correct Answer - D | |
| 10. |
A non-conducting disk of radius `R` is rotating about its own axis with constant angular velocity `omega` in a perpendicular uniform magnetic field `B` as shown in figure. The emf induced between centre and rim of disk is A. `(B omega R^(2))/(2)`B. `B omega R^(2)`C. ZeroD. `(B omega R^(2))/(3)` |
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Answer» Correct Answer - C As the ring is non-conducting, no motion of charge carries occur and hence there is no induced emf |
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| 11. |
A conducting rod `PQ` of length `l=1.0m` is moving with a uniform speed `v2.0m//s` in a uniform magnetic field `B=4.0T` directed into the paper. A capacitor of capacity `C=10muF` is connected as shown in figure. Then |
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Answer» The motional emf is `:. P.d` across the capacitor `= Blv = 4 xx 1 xx 2 = 8V` `q = CV = 10 xx 8 = 80 muC` `A` is `+Ve` w.r.t `B` (From fleming right hand rule) The charge on plate `A` is `q_(A) = 80 muC` The charge on plate `B` is `q_(B) = -80 muC` |
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| 12. |
When a certain circuit consisting of a constant e.m.f. E an inductance L and a resistance R is closed, the current in, it increases with time according to curve 1. After one parameter ( E , L or R ) is changed, the increase in current follows curve 2 when the circuit is closed second time. Which parameter was changed and in what direction A. L is increasedB. L is decreasedC. R is increasedD. R is decreased |
| Answer» Correct Answer - A | |
| 13. |
`AB` and `CD` are fixed conducting smooth rails placed in a vertical palne and joined by a constant current source at its upper end. `PQ` is a conducting rod which is free to slide on the rails. A horizontal uniform magnetic field exists in space as shwn in figure. If the rod `PQ` is released from rest then, A. the rod `PQ` will move downward with constant accelerationB. the rod `PQ` will move upward with constant accelerationC. the rod will remain at restD. any of the above |
| Answer» Correct Answer - D | |
| 14. |
An inductance stroes energy in theA. eelctric fieldB. magnetic fieldC. resistance of the coilD. electric and magnetic fields |
| Answer» Correct Answer - B | |
| 15. |
Three idential coils `A,B` and `C` carrying currents are placed coaxially with their planes parallel to one another. `A` and `C` carry current as shown in figure `B` is kept fixed while `A` and `C` both are moved towards `B` with the same speed. Initially, `B` is equally separated from `A` and `C`. The direction of the induced current in the coil `B` is A. same as that in coil `A`B. same as that in coil `B`C. zeroD. none of these |
| Answer» Correct Answer - C | |
| 16. |
An LCR curcuit is equivalent to a damped pendulum. In an LCR circuit the capacitor is charged to `(Q_0)` and then connected to the L ans R as shown below. If a student plaots graphs of the square of maximum charge `(Q_(max)^(2))` on the capacitor with time (t) for two different values `L_(1) and L_(2) (L_(1)gtL_(2))` of L then which of the following represents this graph correclty? (plots are schematic and not drawn to scale).A. B. C. D. |
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Answer» Correct Answer - A For a demped pendulum, `A = A_(0e^(-bt//2m))` `rArr A = A_(0)e^(-((R )/(2L))t)` (Since `L` plays `s` the same role as `m`) |
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| 17. |
A thin circular ring of area A is held perpendicular to a uniform magnetic field of induction B. A small cut is made in the ring and a galvanometer is connected across the ends such that the total resistance of the circuit is R. When the ring is suddenly squeezed to zero area, the charge flowing through the galvanometer isA. `(BR)/(A)`B. `(AB)/(R )`C. `ABR`D. `(B^(2)A)/(R^(2))` |
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Answer» Correct Answer - B `Q = (Delta phi)/(R ) - (phi_(2)-phi_(1))/(R ) = (BA-0)/(R ) = (BA)/(R )` `= 6 xx 10^(-3) = 6m Wb` |
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| 18. |
An alternating current `I` in an inductance coil varies with time `t` according to the graph as shown: Which one of the following graph gives the variation of voltage with time? A. B. C. D. |
| Answer» Correct Answer - C | |
| 19. |
A rectangular loop is being pulled at a constant speed `v`, through a region of certain thickness `d`, in which a uniform magnetic field `B` is set up. The graph between position `x` of the right hand edge of the loop and the induced e.m.f. `E` will be A. B. C. D. |
| Answer» Correct Answer - B | |
| 20. |
shows a conducting loop being pulled out of a magnetic field with a speed v. Which of lthe four plots shown in may represent the power delivered by the pulling agent as a function of the speed v ? A. aB. bC. cD. d |
| Answer» Correct Answer - B | |
| 21. |
A small square loop of wire of side l is placed inside a large square loop of wire of side `L(Lgtgtl)`. The loops are co-planer and their centres coincide. The mutual inductance of the system is proportional toA. `L//l`B. `l//L`C. `L^(2)//l`D. `l^(2)//L` |
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Answer» Correct Answer - D Flux `= Mi = BA, M prop (l^(2))/(L)` |
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| 22. |
A small square loop of wire of side `l` is placed inside a large square loop of wire of side `L (L gt gt l)`. The loops are coplanar and their centre coincide. What is the mutual inductance of the system ?A. `l//L`B. `l^(2)//L`C. `L//l`D. `L^(2)//l` |
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Answer» Correct Answer - B Magnetic field produced by a current`i` in a large square loop at its centre `B prop (i)/(L)" "` say `B = K(i)/(L)` `:.` Magnetic flux linked with smaller loop, `phi = B.S " " phi = (K(i)/(L))(l^(2))` Therefore, the mutual inductance `M = (phi)/(i) = K(l^(2))/(L)` or `M prop (l^(2))/(L)` |
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| 23. |
A rectangular loop with a sliding connector of length `l = 1.0 m` is situated in a uniform magnetic field `B = 2T` perpendicular to the plane of loop. Resistance of connector is `r = 2 Omega`. Two resistances of `6Omega` and `3Omega` are connected as shown in . The external force required to keep the connetor moving with a constant velocity `v = 2 m s^(-1)` is A. `6N`B. `4N`C. `2N`D. `1N` |
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Answer» Correct Answer - B `F = (B^(2)l^(2)v)/(R_(P))` here `R_(P) = (6 xx 3)/(6+3) = 2 Omega` |
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| 24. |
A specially uniform magnetic field of `0.080 T` is directed inot the plane of the page and perpendicular to it, as shown in the accompanying figure. A wire loop in the plane of the page has constant area `0.010 m^(2)`. The magnitude of the magnetic field decreases at a constant rate of `3.0 xx 10^(-4) T//s`. The magnitude and direction of the induced emf is A. `3.0 xx 10^(-6) V` clockwiseB. `3.0 xx 10^(-6)V` anticlockwiseC. `2.4 xx 10^(-5) V` anticlockwiseD. `8.0 xx 10^(-4)V` clockwise |
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Answer» Correct Answer - A `e = A(dB)/(dt) = 10^(-2) xx 3 xx 10^(-4) = 3 xx 10^(-6)` |
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| 25. |
A magnetic field directed into the page changes with time according to `B = (0.0300t^(2) + 1440)T`, where `t` is in seconds. The field has a circular cross section of radius `R = 2.50 cm`. What are the magitude and direction of the electric field at point `P_(1)` when `t = 3.00 s` and `r_(1) = 0.0200 m`? |
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Answer» `e = oint E.dl = (+d phi)/(dt)` `E = (2pir) = A.(dB)/(dt) = pi r^(2) xx (d)/(d t)(0.03t^(2)+1.4)` `E = (pi r^(2))/(2pi r) xx (0.06 t) = (r )/(2)(0.06 t)` `|E| = (0.02)/(2) xx 0.06 xx 3 = 18 xx 10^(-4) N//C` |
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| 26. |
Whenever the magnet flux linked with a coil changes, then is an induced emf in the circuit. This emf lastsA. For a short timeB. For a long timeC. For everD. So long as the change in the flux takes place |
| Answer» Correct Answer - D | |
| 27. |
The ratio of secondry to the primary turns in a trans is `9:4`. If the power input be `P`, what will be the ratio of power output (neglect all losses) to power inputA. `4:9`B. `9:4`C. `5:4`D. `1:1` |
| Answer» Correct Answer - D | |
| 28. |
The wing span of an aeroplane is `20` metre. It is flying in a field, where the verticle component of magnetic field of earth is `5xx10^(-5)` tesla, with velocity `360 km//h`. The potential difference produced between the blades will beA. `0.10 V`B. `0.15 V`C. `0.20 V`D. `0.30 V` |
| Answer» Correct Answer - A | |
| 29. |
The number of turns in the coil of an ac genrator is `5000` and the area of the coil is `0.25m^(2)`. The coil is rotate at the rate of `100 "cycles"//"sec"` in a magnetic field of `0.2W//m^(2)`. The peak value of the emf generated is nearlyA. 786 kVB. 440 kVC. 220 kVD. 15.71 kV |
| Answer» Correct Answer - D | |
| 30. |
The armature of dc motor has `20Omega` resistance. It draws current of `1.5` ampere when run by `220` volts dc supply. the value of back e.m.f. induced in it will beA. 150 VB. 170 VC. 180 VD. 190 V |
| Answer» Correct Answer - D | |
| 31. |
Assertion : An electric motor will have maximum efficiency when back emf becomes equal to half of applied emf. Reason : Efficiency of electric motor depends only on magnitude of back emf.A. If both assertion and reason are t rue and the reason is the correct explanation of the assertion.B. If both assertion and reason are true but reason is not the correct explanation of the assertionC. If assertion is true but reason is false.D. If the assertion and reason both are fal se. |
| Answer» Correct Answer - D | |
| 32. |
A bar of mass `m` and length `l` moves on two frictionless parallel rails in the presence of a uniform magnetic field directed into the plane of the paper. The bar is given and initial velocity `v_(i)` to the right and released. Find the velocity of bar, induced emf across the bar and the current in the circuit as a function of time |
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Answer» The induced current in the counter clockwise direction and the magnetic force on the bar is given by `F_(B) = -ilB`. The negative sign indicates that the force is towards the left and retards motion. `F = m a` `-i lB = m.(dv)/(dt)` Because the force depends on current and the current depends on the speed, the force is not constant and the acceleration of the bar is not constant. The induced current is given by `i = (B l v)/(R )`, `-ilB = m.(d v)/(d t)` `-((Blv)/(R ))lB = m.(dv)/(dt) rArr (dv)/(dt) = -(B^(2)l^(2))/(mR)dt` `underset(v_(1))overset(v)int(dv)/(v) = -(B^(2)l^(2))/(mR)underset(0)overset(t)intdt`, `ln ((v)/(v_(1))) = -(B^(2)l^(2))/(mR)t = (-t)/(T)` where `T = (mR)/(B^(2)l^(2)) rArr v = v_(i)e^((t)/(T))` The speed of the bar therefore decreases exponentially with time under the action of magnetic retarding force. `emf = iR = Blv_(i)e^((t)/(T))`, current: `i = (Blv)/(R ) = (B l)/(R )v_(1)e^((t)/(T))` |
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| 33. |
In an oscillating LC circuit the maximum charge on the capacitor is Q. The charges on the capacitor when the energy is stored equally between the electric and magnetic field isA. `Q`B. `(Q)/(2)`C. `(Q)/(sqrt(3))`D. `(Q)/(sqrt(2))` |
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Answer» Correct Answer - D `(1)/(2)(Q^(2))/(2C) = (q^(2))/(2C)` `rArr = q = (Q)/(sqrt(2))` |
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| 34. |
Two coaxial solenoids are made by winding thin insulated wire over a pipe of cross-sectional area `A=10 cm^(2)` and length =20cm. If one of the solenoid has 300 turns and the other 400 turns, their mutual indcutance is |
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Answer» `M = (mu_(0)N_(1)N_(2)A)/(L)` `M = (4pi xx 10^(-7) xx 3 xx 10^(2) xx 4 xx 10^(2) xx 10^(-3))/(2 xx 10^(-1))` `= 2.4 pi xx 10^(-4)H` |
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| 35. |
Two coaxial solenoids are made by winding thin insulated wire over a pipe of cross-sectional area `A=10 cm^(2)` and length =20cm. If one of the solenoid has 300 turns and the other 400 turns, their mutual indcutance isA. `4.78 pi xx 10^(-5)H`B. `2.4 pixx 10^(-4) H`C. `2.4 pixx 10^(-5) H`D. `4.8pi xx 10^(-4) H` |
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Answer» Correct Answer - B `M = mu_(0) n_(1) n_(2) pi r_(1)^(2) l` From `phi_(2) = pi_(1)^(2)(mu_(0) n i)n_(21)` `A = pi r_(1)^(2) = 10 cm^(2), l = 20 cm, N_(1) = 300`, `N_(2) = 400` `M = (mu_(0) N_(1)N_(2)A)/(l)` `(4pi xx 10^(-7) xx 300 xx 400 xx 10 xx 10^(-4))/(0.20)` `= 2.4 pi xx 10^(-4) H` |
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| 36. |
If a change in current of `0.01 A` in one coil produces change in magnetic flux of `1.2xx10^(-2)Wb` in the other coil, then the mutual inductance of the two coils in henries |
| Answer» Correct Answer - C | |
| 37. |
A coil of `100` turns with a current of `5A` produce a magnetic flux of `1 mu Wb` and each turn of the coil. The coefficeint of self induction isA. `10 mu H`B. `20 mu H`C. `30 mu H`D. `40 mu H` |
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Answer» Correct Answer - B `n phi = L i` |
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| 38. |
In a uniform magneitc field of induced B a wire in the form of a semicircle of radius r rotates about the diameter of hte circle with an angular frequency `omega`. The axis of rotation is perpendicular to hte field. If the total resistance of hte circuit is R, the mean power generated per period of rotation isA. `(B pi r^(2) omega)/(2R)`B. `((B pi r^(2) omega)^(2))/(8R)`C. `((B pi r omega)^(2))/(2R)`D. `((B pi r omega H_(2))^(2))/(8R)` |
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Answer» Correct Answer - B `phi = BA cos theta` `= B((pi r^(2))/(2)) cos omega t` `E = -(d)/(dt) phi = (B pi r^(2))/(2) omega sin omega t` `P = (E^(2))/(R ) = (B^(2) pi^(2)r^(4)omega^(2))/(4R) sin^(2) omega t` But `lt sin^(2) omega t gt (1)/(2)` `:. lt P gt =((B pi r^(2) omega)^(2))/(8 R)` |
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| 39. |
An e.m.f. of `5 "volt"` is produced by a self-inductance, when the current changes at a steady rate from `3 A` to `2A` `1`millisecond. The value of self-inductance isA. ZeroB. 5 HC. 5000 HD. 5mH |
| Answer» Correct Answer - D | |
| 40. |
A coil has self inductance of `0.01 H`. The current through it is allowed to change at the rate of `1A` in `10^(-2)s`. The induced emf isA. `1V`B. `2V`C. `3V`D. `4V` |
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Answer» Correct Answer - A `e = L(di)/(dt)` |
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| 41. |
A coil has `2000` turns and area of `70cm^(2)`. The magnetic field perpendicular to the plane of the coil is `0.3 Wb//m^(2)` and takes `0.1` sec to rotate through `180^(0)`. The value of the induced e.m.f. will beA. `8.4 V`B. `84 V`C. `42 V`D. `4.2 V` |
| Answer» Correct Answer - B | |
| 42. |
A varying current in a coil change from `10A` to `0A` in `0.5 sec`. If the average emf induced in the coil is `220V`, the self inductance of the coil isA. 5HB. 10 HC. 11HD. 12H |
| Answer» Correct Answer - C | |
| 43. |
A magnetic field of `2xx10^(-2)T` acts at right angles to a coil of area `100cm^(2)` with `50` turns. The average emf induced in the coil is `0.1V`, when it is removed from the field in time `t`. The value of `t` isA. `0.1 sec`B. `0.01 sec`C. `1 sec`D. `20 sec` |
| Answer» Correct Answer - A | |
| 44. |
A coil of N turns and mean cross-sectional area A is rotating with uniform angular velocity `omega` about an axis at right angle to uniform magnetic field B. The induced emf E in the coil will beA. `NBA sin omega t`B. `NB omega sin omega t`C. `NB//A sin omega t`D. `NBA omega sin omega t` |
| Answer» Correct Answer - D | |
| 45. |
A coil having n turns and resistance `R Omega` is connected with a galvanometer of resistance `4R Omega`. This combination is moved in time t seconds from a magnetic field `W_(1)` weber to `W_(2)` weber. The induced current in the circuit isA. `-(W_(2)-W_(1))/(5 Rnt)`B. `-(n(W_(2)-W_(1)))/(5 Rnt)`C. `-((W_(2)-W_(1)))/(Rnt)`D. `-(n(W_(2)-W_(1)))/(Rt)` |
| Answer» Correct Answer - B | |
| 46. |
A coil of `40 Omega` resistance has `100` turns and radius `6 mm` us connected to ammeter of resistance of `160 ohms`. Coil is placed perpendicular to the magnetic field. When coil is taken out of the field, `32 mu C` charge flows through it. The intensity of magnetic field will beA. `6.55 T`B. `5.66 T`C. `0.655 T`D. `0.566 T` |
| Answer» Correct Answer - D | |
| 47. |
Magnetic flux in a circuite containing a coil of resistance `2Omega`change from `2.0Wb` to `10 Wb` in `0.2 sec`. The charge passed through the coil in this time isA. `0.8 C`B. `1.0 C`C. `5.0C`D. `4.0 C` |
| Answer» Correct Answer - D | |
| 48. |
Magnetic flux in a circuite containing a coil of resistance `2Omega`change from `2.0Wb` to `10 Wb` in `0.2 sec`. The charge passed through the coil in this time isA. 5.0 coulombB. 4.0 coulombC. 1.0 coulombD. 0.8 coulomb |
| Answer» Correct Answer - B | |
| 49. |
In the circuit shown in Fig. switch `k_(2)` is open and switch `k_(1)` is closed at `t = 0`. At time `t = t_(0)`, switch `k_(1)` is open and switch `k_(2)` is simultaneosuly closed. The variation of inductor current with time is A. B. C. D. |
| Answer» Correct Answer - A | |
| 50. |
Consider a magnet surround by a wire with an on`//` of switch `S` (figure) if the switch is thrown from the off position (open circuit) to the on position the (closed circuit). A. The current will flow in `C.W` direction as seen from the left sideB. the current will flow in anti clock wise direction as seen from `L.H.S`C. The current will not flowD. Can not be decided. |
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Answer» Correct Answer - C When the switch is thrown the off position (open circuit) to the position (closed circuit) then neither `B`, nor `A` the angle between `B` and `A` change thus no change in magnetic flux linked with coil occur, hence no electromotive force is produced and consequency no current will flow in the circuit. |
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