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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. |
A current `i` ampere flows in a circular arc of wire whose radius is `R`, which subtend an angle `3pi//2` radian at its centre. The magnetic induction `B` at the centre is A. `(mu_(0)i)/R`B. `(mu_(0)i)/(2R)`C. `(2mu_(0)i)/R`D. `(3 mu_(0)i)/(8R)` |
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Answer» Correct Answer - D `B=(mu_(0))/(4pi)((2pi- theta)i)/R=(mu_(0))/(4pi)((2pi-pi/2)xxi)/R=(3mu_(0)i)/(8R)` |
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| 52. |
An `alpha-`particle is describing a circle of radius 0.45 m in a field of magnetic induction of 1.2 T. Find its speed, frequency of rotation and kinetic energy. What potential difference will be required to accelerate the particle so as to give this much of the energy to it? The mass of `alpha-`particle is `6.8 xx 10^(-27) kg` and its charge is twice the charge of proton, i.e.,`3.2 xx 10^(-19) C`. |
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Answer» Correct Answer - `2.6xx10^7 ms^-1 ; 9.2xx10^6s^-1 ; 14.3 MeV ; 7xx10^6V`. we know that `F=evB=(mv^2)/r or v=(eBr)/m` Subtituting the given values, we get `v=((3.2xx10^-19)(1.2)(0.45))/((6.8xx10^-27))=2.6xx10^7ms^-1` The frequency of rotation is given by `f=v/(2pir)=(2.6xx10^7)/(2xx3.14xx0.45)=9.2xx10^6s^-1` The kinetic energy of `alpha`-particle is given by `K=1/2mv^2=1/26.8xx10^-27xx(2.6xx10^7)^2=2.3xx10^-12J` `=(2.3xx10^-12)/(1.6xx10^-19)eV=14.3xx10^6eV=14.3MeV` |
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| 53. |
A deutron of kinetic energy `50` keV is describing a circular orbit of radius `0.5` meter in a plane perpendicular to magnetic field `vecB`. The kinetic energy of the proton that describes a circular orbit of radius `0.5` meter in the same plane with the same `vecB` isA. 25 keVB. 50 keVC. 200 keVD. 100 keV |
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Answer» Correct Answer - D |
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| 54. |
A proton, a deutron and `alpha`-particle, whose kinetic energies are same, enter perpendicularly a uniform magnetic field. Compare the radii of their circualr paths.A. the radius `r_(d)` of the deutron path to the radius `r_(p)` of the proton path is `sqrt2`B. the radius `r_(alpha)` of the alpha particle path to `r_(p)` is 1C. the radius `r_(d)` of the deutron path to the radius `r_(p)` of the proton path is 2D. the radius `r_(alpha)` of the alpha particle path to `r_(p)` is `sqrt2` |
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Answer» Correct Answer - A::B |
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| 55. |
A proton , a deutron and `alpha )`- particle having the same kinetic energy are moving in circular trajectories in a constant magnetic field . If ` r_(p), r_(d) , and r_(alpha)` denote respectively the radii of the trajectories of these particles , thenA. `r_(alpha)=r_(p) lt r_(d)`B. `r_(alpha) gt r_(d) lt r_(p)`C. `r_(alpha)=r_(d) gt r_(p)`D. `r_(p) = r_(d) = r_(alpha)` |
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Answer» Correct Answer - A Given that `K_(p) - K_(d) = K_(a) = K` (say) We know that `m_(p0 = m, m_(d) = 2m` and `m_(a) = 4m` and `q_(p) = e, q_(d) = 2e` Further`r = (sqrt(2mK))/(qB) implies r_(p) = (sqrt(2mK))/(eB)`. `r_(d) = (sqrt(2 (2m) K))/(eB) = sqrt(2) r_(p)` and `r_(a) = (sqrt(2(4m) K))/((2e) B) = r_(p)`. Hence `r_(alpha) = r_(p) lt r_(d)` |
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| 56. |
Assertion: If two long wires, hanging freely are connected to a battery in series, they come closer to each other. Reason: Force of attraction acts between the two wires carrying currents.A. If both the assertion and reason are true and reason is a true explantion of the assertion.B. If both the assertion and reason are true but the reason is not true the correct explantion of the assertion.C. If the assertion is true but reason falseD. If both the assertion and reason are false. |
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Answer» Correct Answer - D When two long parallel wires, are connected to a battery in series. They carry currents in opposite directions, hence they repel each other. |
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| 57. |
Three long, straight parallel wires carrying current, are arranged as shown in figure. The force experienced by a `25cm` length of wire `C` is A. `10^(-3)`B. `2.5xx10^(-3)` NC. ZeroD. `1.5xx10^(-3)` N |
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Answer» Correct Answer - C |
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| 58. |
A wire of 60 cm length and mass 16 gm is suspended by a pair of flexible leads in a magnetic field of induction 0.40 T. What are the magnitude and direction of the current required to remove the tension in the supporting leads? |
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Answer» We know that the magnetic force on a wire is given by `F=ilB sin theta` Here B is at right angle to I and hence `theta=90^@`. So, `F=ilB` Setting `F=mg`, the weight of the wire, the magnitude of the current required to remove the tension in the supporting lead is `i=(mg)/(lB)=((1.0xx10^-2kg)(10ms^-2))/((0.6m)(0.4Wbm^-2))=0.41A` Since the magnetic force should act in upward direction so the direction of current should be from left no right. |
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| 59. |
Three long straight wires are connected parallel to each other across a battery of negligible internal resistance. The ratio of their resistances are `3:4:5`. What is the ratio of distances of middle wire from the others if the net forces experienced by it is zero.A. `4:3`B. `3:1`C. `5:3`D. `2:3` |
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Answer» Correct Answer - C |
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| 60. |
A metal rod of mass 10 gm and length 25 cm is suspended on two springs as shown in Fig. 1.131. The springs are extended by 4 cm. When a 20 A current passes through the rod, it rises by 1 cm. Determine the magnetic field assuming acceleration due to gravity to be `10 ms^-1.` |
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Answer» Correct Answer - `1.5xx10^-3T` Initially, the rod will be in equilibrium if `2T_0=mg with T_0=kx_0....(i)` When current is passed through the rod, it experiences a force `F=Bil` vertically upward. Now for equilibrium, in this situation `2T+Bil=mg with T=kx......(ii)` From equations (i) and (ii), we get `T/(T_0)=(mg-Bil)/(mg) or x/(x_0)=1-(Bil)/(mg)` Solve to get `B=(mg(x_0-x))/(ilx_0)=((1xx10^-3)(10)(3xx10^-2))/((20)(25xx10^-2)(4xx10^-2))` `=1.5xx10^-3T` |
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| 61. |
Two metal strips, each of length, each of length l, are clamped parallel to each other on a horizontal floor with a separation b between them. A wire of mass m line on them perpendicularly as shown in figure. A vertically upward magnetic field of strenght B exists in the space. The matal strips are smooth but the cofficient of freiction between the wire and the floor is `mu`. A current i is established when the switch S is closed at the instant `t=0`. Discuss the motion of the wire after the switch is closed. How far away from the strips will the wire reach? |
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Answer» Correct Answer - `(IBbl)/(mu mg)` When the switch is closed, current flows in the wire. Due to this current, magnetic force acts on wire towards right. The acceleration of wire : `a=(F_(b))/(m)=(IBb)/(m)` Velocity gained by wire till it reaches the end of rails: `v^2=2al=(2IBbl)/m` After this, let wire slips a distance s on the floor `mumgs=1/2mv^2 implies mumgs=IBbl implies s=(IBbl)/(mumg)` |
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| 62. |
Two concentric coplanar circular loops of radii `r_(1)` and `r_(2)` carry currents of respectively `i_(1)` and `i_(2)` in opposite direction (one clockwise and the other anticlockwise). The magnetic induction at the centre of the loops is half that due to `i_(1)` alone at the centre. if `r_(2)=2r_(1)`. the value of `i_(2)//i_(1)` isA. `2`B. `1//2`C. `1//4`D. `1` |
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Answer» Correct Answer - D Magnetic field at centre due to smaller loop `B_(1)=(mu_(0))/(4pi). (2pii_(1))/(r_(1)).......(i)` Due to Bigger loop `B_(2)=(mu_(0))/(4pi). (2pii_(2))/(r_(2)).......(i)` So net magnetic field at centre `B=B_(1)-B_(2)=(mu_(0))/(4pi)xx2pi((i_(1))/(r_(1))-(i_(2))/(r_(2)))` According to question `B=1/2xxB_(1)` `(mu_(0))/(2pi).2pi((i_(1))/(r_(1))-(i_(2))/(r_(2)))=1/2xx(mu_(0))/(4pi). (2pii_(1))/(r_(1))` `(i_(1))/(r_(1))-(i_(2))/(r_(2))=(i_(1))/(2r_(1)) implies (i_(1))/(2r_(1))=(i_(2))/(r_(2)) implies (i_(2))/(r_(2))=1{r_(2)=2r_(1)}` |
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| 63. |
A straight wire lies along a body diagonal of an imaginary cube of side `a=20 cm,` and carries a current of 5 A (as shown in Fig. 1.61). Find the force on it due to a uniform field `vec B = 0.6 hat j T.` |
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Answer» In order to use equation `vecF=iveclxxvecB`, we would need to find the angle between `vecl and vecB`. We avoid this task by using unit vector notation. Form the problem figure, we see that `vecl=ahati-ahatj+ahatk` Thus, the force is `vecF=IveclxxvecB=IaB(hati-hatj+hatk)xx(hatj)` `=IaB(-hati+hatk)` The force lies in the xz plane and has a magnitude `F=sqrt2IaB=0.85N` |
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| 64. |
A current of `10` ampere is flowing in a wire of length `1.5m`. A force of `15 N` acts on it when it is placed in a uniform magnetic field of `2` tesla. The angle between the magnetic field and the direction of the current isA. `30^(@)`B. `45^(@)`C. `60^(@)`D. `90^(@)` |
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Answer» Correct Answer - A `F=Bil sinthetaimpliessintheta=(F)/(Bil)=(15)/(2xx10xx1.5)=(1)/(2)` ` impliestheta=30^(@)` |
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| 65. |
Due to `10` ampere of current flowing in a circular coil of `10 cm` radius, the magnetic field produced at its centre is `3.14xx10^(-3) Weber//m^(3)`. The number of turns in the coil will beA. `5000`B. `100`C. `50`D. `25` |
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Answer» Correct Answer - C `B=(mu_(0))/(4pi).(2piNi)/r` `implies 3.14x10^(-3)=(10^(-7)xx2xx3.14xxNxx10)/((10xx10^(-2)) implies N=50` |
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| 66. |
In Fig. the bar AC has a mass of 50 g. It slides frictionlessly on the metal strips 40 cm apart at the edges of the incline. A current I flows through these strips and the bar, as shown. There is a magnetic field `|B_x| = 0.02 T` directed in the -y direction. How much must I be if the rod is to remain motionless? Neglect the slight overhang of the rod. |
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Answer» Magnetic force is along the positive x-axis. If motion is to occur along the incline, `sumF=0:IL|B_y|cos37^@=mgsin37^@` from which `I=(mg tan37^@)/(L|B_y|)=((0.050)(9.8)(0.75))/((0.40)(0.02))=46A` |
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| 67. |
`A` and `B` are two concentric circular conductors of centre `O` and carrying currents `i_(1)` and `i_(2)` as shown in the adjacent figure. If ratio of their radii is `1:2` and ratio of the flux densities at `O` due to `A` and `B` is `1:3`, then the value of `i_(1)//i_(2)` is A. `1/6`B. `1/4`C. `1/3`D. `1/2` |
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Answer» Correct Answer - A `r_(1):r_(2)=1:2` and `B_(1):B_(2)=1:3` we know that `B=(mu_(0))/(4pi).(2pini)/r implies (i_(1))/(i_(2))=(B_(1)r_(1))/(B_(2)r_(2))=(1xx1)/(3xx2)=1/6` |
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| 68. |
A conductor bar with mass m and length L slides over horizontal rails that are connected to a voltage source. The voltage source maintains a constant current I in the rails and bar, and constant, uniform, vertical magnetic field `vecB` fills the region between the rails (as shown in Fig.) Find the magnitude and direction of the net force on the conducting bar. Ignore friction, air resistance and electrical resistance.A. ILB, to the right.B. ILB, to the left.C. 2ILB, to the right.D. 2ILB, to the left. |
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Answer» Correct Answer - a `F=ILB`, to the right. `v^2=2ad implies d=(v^2)/(2a)=(v^2m)/(2ILB)` |
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| 69. |
A conductor bar with mass m and length L slides over horizontal rails that are connected to a voltage source. The voltage source maintains a constant current I in the rails and bar, and constant, uniform, vertical magnetic field `vecB` fills the region between the rails (as shown in Fig.) If the bar has mass m, find the distance d that the bar must move along the rails from rest to attain speed v.A. `(3v^2m)/(2ILB)`B. `(5v^2m)/(2ILB)`C. `(v^2m)/(ILB)`D. `(v^2m)/(2ILB)` |
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Answer» Correct Answer - d `F=ILB`, to the right. `v^2=2ad implies d=(v^2)/(2a)=(v^2m)/(2ILB)` |
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| 70. |
Two long conductors, separated by a distance `d` carry current `I_(1) and I_(2)` in the same direction . They exert a force `F` on each other. Now the current in one of them is increased to two times and its direction is reversed . The distance is also increased to `3d`. The new value of the force between them isA. `-2F`B. `(F)//(3)`C. `(2F)//(3)`D. `(-F)//(3)` |
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Answer» Correct Answer - C `Fprop(i_(1)i_(2))/(a)`, Since one of the current increase two times and distance increases three times, so force become `(2)/(3)` times. Also due to the reversal of direction current force becomes negative. |
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| 71. |
`A, B` and `C` are parallel conductors of equal length carrying currents `I, I` and `2I` respectively. Distance between `A` and `B` is `x`. Distance between `B` and `C` is also `x`. `F_(1)` is the force exerted by `B` on `A` and `F_(2)` is the force exerted by `B` on `A` choose the correct answer A. `F_(1)=2F_(2)`B. `F_(2)=2F_(1)`C. `F_(1)=F_(2)`D. `F_(1)=-F_(2)` |
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Answer» Correct Answer - D `F=(mu_(0))/(4pi)(2i_(1)i_(2))/(a)` `F_(1)=(mu_(0))/(4pi)=(2i^(2))/(x)` (Attraction) `F_(2)=(mu_(0))/(4pi)=(2ixx2i)/(2x)=(mu_(0))/(4pi)(2i^(2))/(x)` (Repulsion) Thus `F_(1)=-F_(2)` |
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| 72. |
Current `I_1 and I_2` flow in the wires shown in Fig. The field is zero at distance x to the right of O. Then A. `x=((I_(1))/(I_(2)))a`B. `x=((I_(2))/(I_(1)))a`C. `x=((I_(1)-I_(2))/(I_(1)+I_(2)))a`D. `x=((I_(1)+I_(2))/(I_(1)-I_(2)))a` |
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Answer» Correct Answer - C `(mu_(0)I_(1))/(4pi(a+x))=(mu_(0)I_(2))/(4pi(a-x))` `(a-x)/(a+x)=(I_(2))/(I_(1))` `I_(1)a-I_(1)x=I_(2)a-I_(2)x` `x=((I_(1)-I_(2))/(I_(1)+I_(2)))a` |
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| 73. |
A current of `1//(4pi)` ampere is flowing in a long straight conductor. The line integral of magnetic induction around a closed path enclosing the current carrying conductor isA. `10^(-7)` weber per meterB. `4pixx10^(-7)` weber per meterC. `16pi^(2)xx10^(-7)` weber per meterD. zero |
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Answer» Correct Answer - A `ointvecB.vecdvecI=mu_(0)I` `=4pixx10^(-7)xx1/(4pi)` `=10^(-7) Wb//m` |
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| 74. |
A metal wire PQ of mass 10g lies at rest on two horizontal metal rails separated by 5cm as shown in Fig. A vertically downward magnetic field of magnitude 0.80T exist in the space. The resistance of the circuit is slowly decreased and it is found that when the resistance goes below `20.0Omega` the wire PQ starts sliding on the rails. The coefficient of friction between wires and rails is found. The coefficient of friction between wires and rails is found to be `n//25`. find n |
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Answer» Correct Answer - 3 `Ibl=mumg` `implies 6/20xx0.8xx5/100=mu10/1000(10)` `implies mu=0.12=3//25` |
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| 75. |
A unifrom conducting wire `ABC` has a mass of `10g`. A current of `2 A` flows through it. The wire is kept in a unifrom magnetic field `B = 2T`. The accleration of the wire will be A. ZeroB. `12 ms^(-2)` along y-axisC. `1.2xx10^(-3) ms^(-2)` along y-axisD. `0.6xx10^(-3) ms^(-2)` along y-axis |
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Answer» Correct Answer - B |
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| 76. |
A thin, 50 cm long metal bar with mass 750 g rests on, but is not attrached to, two metallic supports in a uniform 0.450T magnetic field, as shown in fig. A battery and a `25Omega` resistor in series are connected to the supports. What is the largest voltage the battery can have without breaking the circuit at the supports?A. `817V`B. `412V`C. `325V`D. `160V` |
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Answer» Correct Answer - a (a) `F_1=mg` When bar is just ready to levitate, `IlB=mg, I=(mg)/(lB)=((0.75kg)(9.80m//s^2))/((0.500m)(0.450T))=32.67T` `epsilon=IR=(32.67A)(25.0Omega)=817V` (b) `R=2.0Omega, I=epsilon/R=(817V)(2.0Omega)=408.5A` `F_1=IlB=92N` `a=(F_1-mg)//a=113ms^-2` |
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| 77. |
A copper rod of mass m rests on two horizontal rails distance L apart and carries a current of I from one rail to the other. The coefficient of static friction between rod and rails is `mu_(s)` What are the (a) magnitude and (b) angle (relative to the vertical) of the smallest magnetic field that puts the rod on the verge of sliding? |
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Answer» Correct Answer - `(a)0.10T(b)31^(@)` |
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| 78. |
A long, straight, hollow conductor (tube) carrying a current has two sections A and C of unequal cross sections joined by a conical section B. 1,2 and 3 are points on a line parallel to the axis of the conductor. The magnetic fields at 1,2 and 3 have magnitudes `B_1, B_2 and B_3`. Then, A. `B_(1)=B_(2)=B_(3)`B. `B_(1)=B_(2)neB_(3)`C. `B_(1) lt B_(2) lt B_(3)`D. `B_(2)` cannot be found unless the dimensions of the section `B` are known |
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Answer» Correct Answer - A To find the magnetic field outside a thick conductor, the current may be assumed to flow along the axis. As point `1,2,3` are equidistance from the axis. `B_(1)=B_(2)=B_(3)` |
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| 79. |
Two long straight parallel conductors are separated by a distance of 5 cm and carrying current 20 A. what work per unit length of a conductor must be done to increases the separation between conductors to 10 cm, if the current flows in the same direction ?A. `8xx10^(-5) log_(e)2`B. `log_(e)2`C. `10^(-7) log_(e) 2`D. none of these |
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Answer» Correct Answer - a The force per unit length is `F_(0)=(mu_(0)l_(1)l_(2))/(2pi) int_(5xx10^(-2))^(10xx10^(-2)) (dr)/r` `=(mu_(0)l_(1)l_(2))/(2pi) ln 2 =2xx10^(-7) xx20xx20 ln 2` `=8xx10^(-5) ln 2=8xx10^(-5) log_(e) 2` |
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| 80. |
A long straight conductor carrying current `I_1` is placed in the plane of a ribbon carrying current `I_2` parallel to the previous one. The width of the ribbon is b and the straight conductor is at a distance a from the near endge. Find the force of attraction between the two. A. `(mu_(0)I_(1)I_(2))/(2pib)log((a+b)/(a))`B. `(mu_(0)I_(1)I_(2))/(2pib)log((a)/(a+b))`C. `(mu_(0)I_(1)I_(2))/(2pib)log((a)/(b))`D. `(mu_(0)I_(1)I_(2))/(2pib)log(((a+b))/(a))` |
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Answer» Correct Answer - A Consider an elementary strip of width `dx`on the sheet at a distance `x` from the wire. Force on this element is : `dF=(mu_(0)I_(1))/(2pix)I_(2)dx` Total force on unit length of sheet is `F=int_(a)^(a+b)(mu_(0)I_(1))/(2pix)I_(2)dx` `F=(mu_(0)I_(1)I_(2))/(2pi)int_(a)^(a+b)(1)/(x)dx=(mu_(0)I_(1)I_(2))/(2pia)log((a+b)/(b))` |
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| 81. |
Doubly ionized helium ions are projected with a speed of `10kms^(-1)` in a direction perpendicular to a uniformmagnetic field of magnitude 1.0 T. Find (a) the force acting on an ion, (b) the radius of the circle in which it ciruclates and ( c) the time taken by an ion to complete the circle. |
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Answer» (a) `F=qv B sin theta` `=(2xx1.6xx10^-19)xx(10xx10^3)xx1.0xxsin 90^@` `=3.2xx10^-15N` (b) `r=(mv)/(qB)=((4xx1.67xx10^-27)xx10xx10^3)/((2xx1.6xx10^-19)xx1.0)=2.1xx10^-4m` (c) `T=(2pim)/(qB)=(2pixx(4xx1.67xx10^-27))/((2xx1.6xx10^-19)xx1.0)=1.32xx10^-7s` |
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| 82. |
Two identical particles having the same mass m and charges +q and -q separated by a distance d enter a uniform magnetic field B directed perpendicular to paper inwards with speeds `v_1 and v_2` as shown in Fig. 1.139. The particles will not collide if A. `dgt(m)/(Bq)(v_(1)+v_(2))`B. `dlt(m)/(Bq)(v_(1)+v_(2))`C. `dgt(2m)/(Mq)(v_(1)+v_(2))`D. `v_(1)=v_(2)` |
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Answer» Correct Answer - C The particles will not collide if `dgt2(r_(1)+r_(2))` or `dgt2((mv_(1))/(Bq)+(mv_(2))/(Bq))` `dgt(2m)/(Bq) (v_(1)+v_(2))` |
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| 83. |
A horizontal rod of mass `10g` and length `10cm` is placed on a smooth plane inclined at an angle of `60^@` with the horizontal with the length of the rod parallel to the edge of the inclined plane. A uniform magnetic field induction B is applied vertically downwards. If the current through the rod is `1*73ampere`, the value of B for which the rod remains stationary on the inclined plane isA. `1.73` TB. `(1)/(1.73)`TC. 1.TD. None of the above |
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Answer» Correct Answer - C |
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| 84. |
A long, horizontal wire AB rests on the surface of a table and carries a current I. Horizontal wire CD is vertically above wire AB, and is free to slide up and down on the two vetical mental guides C and D (as shown in Fig) Wire CD is connected through the sliding contacts to another wire that also carries a current I, opposite in direction to the current in wire AB. The mass per unit length of the wire CD is `lambda`. To what euqilibrium heigth h will the wire CD rise, assuming that magnetic force on it is wholly due to the current in wire AB? A. `(3mu_(0)I^(2))/(2pilambdag)`B. `(mu_(0)I^(2))/(3pilambdag)`C. `(5mu_(0)I^(2))/(2pilambdag)`D. `(mu_(0)I^(2))/(2pilambdag)` |
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Answer» Correct Answer - D Gravitational force and magnetic force are equal in magnitude and opposite in direction. `implieslambdaLg=(mu_(0)I^(2)L)/(2pih)impliesh=(mu_(0)I^(2))/(2pilambdag)` |
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| 85. |
Two identical particles having the same mass m and charges +q and -q separated by a distance d enter a uniform magnetic field B directed perpendicular to paper inwards with speeds `v_1 and v_2` as shown in Fig. 1.139. The particles will not collide if A. `dgt m/(Bq) (v_1 + v_2)`B. `dlt m/(Bq) (v_1 + v_2)`C. `dgt (2m)/(Bq) (v_1 + v_2)`D. `v_1 = v_2` |
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Answer» Correct Answer - c The particle will not collide if `dgt2(r_1+r_2)` or `dgt2((mv_1)/(Bq)+(mv_2)/(Bq)) or dgt(2m)/(Bq) (v_1+v_2)` |
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| 86. |
Two straight long conductors `AOB` and `COD` are perpendicular to each other and carry currents `I_(1)` and `I_(2)`. The magnitude of the magnetic induction at a point `P` at a distance `d` from the point `o` in a direction perpendicular to the plane `ABCD` is :A. `(mu_(0))/(2pid)(I_(1)+I_(2))`B. `(mu_(0))/(4pid)(I_(1)+I_(2))`C. `(mu_(0))/(2pid)sqrt(I_(2)^(2)+I_(2)^(2))`D. `(mu_(0))/(2pid)((I_(1)I_(2))/(I_(1)+I_(2)))` |
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Answer» Correct Answer - C Here, point `P` is placed symmetrically with respect to two coductors. The magnetic fields due to the conductors are mutually perpendicular. Their resulatant is given by `B_(r)=(mu_(0))/(2pid) sqrt((I_(2)^(2)+I_(2)^(2)))` |
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| 87. |
A conducting ring of mass 2kg and radius `0.5m` is placed on a smooth plane. The ring carries a current of `i=4A`. A horizontal magnetic field `B=10 T` is switched on at time `t=0` as shown in fig The initial angular acceleration of the ring will be .A. `40pirad s^-2`B. `20pirad s^-2`C. `5pirad s^-2`D. `15pirad s^-2` |
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Answer» Correct Answer - a Due to torque of magnetic field, ring will rotate about vertical diameter. `tau=Ialphaimplies MB=Ialpha` `implies ipir^2B=1/2mr^2alpha` `implies alpha=(2iBpi)/m=(2xx4xx10pi)/2=40pi rads^-2` |
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| 88. |
Let current `i=2A` be flowing in each part of a wire frame as shown in Fig. 1.138. The frame is a combination of two equilateral triangles ACD and CDE of side 1 m. It is placed in uniform magnetic field ` B= 4 T` acting perpendicular to the plane of frame. The magnitude of magnetic force acting on the frame is The pithc of the helical path followed by the particle is p. The radius of the helix will beA. 24 NB. zeroC. 16 ND. 8 N |
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Answer» Correct Answer - a `F_(CAD)=F_(CD)=F_(CAD)` `:.` Net force on the frame `=3F_(CD)` `=(3)(2)(1)(4)=24N (F=ilB)`. |
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| 89. |
Two long, parallel conductors carry currents in the same direction, as shown in figure. Conductors A held firmly in position. Conductors B carries a current `I_(B)` and is allowed to slide freely up and down (parallel to A) between a se of non-conducting guides. the mass per unit length of conductors B is 0.1 g/cm and the distance between the two conductors is 5 cm. if system of conductors is in equilibrium , the value of current `I_(B)` is A. 250 AB. 240 AC. 220 AD. 230 A |
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Answer» Correct Answer - a When system of conductors is in equilibrium. The magnetic force of attraction per unit length between conductors =weight of conductors B per unit length. `(mu_(0))/(2pi) (l_(A)xxl_(B))/d=(mg)/L =(m/L)g` `(mu_(0))/(2pi) (l_(A)xxl_(B))/d=(m/L)g` `2xx10^(-7) xx(100xxl_(B))/0.05 =(0.01kg//m) xx10` |
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| 90. |
The rigid conducting thin wire frame carries an electric current I and this frame is inside a uniform magnetic field `vec B` as shown in Fig. 1.146. Then, A. The net magnetic force on the frame is zero but the torque is not zeroB. The net magnetic force on the frame and the torque due to magnetic field are both zero.C. the net magnetic force on the frame is not zero and the torque is also not zero.D. None of these |
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Answer» Correct Answer - b As we know that `vecF=oint Ivec(dl)xxvecB and vectau=vecMxxvecB` |
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| 91. |
Two parallel conductors carry current in opposite direction as shown in figure. One conductor carries a current of 10.0 A. point C is a distance `d/2` to the right of the 10.0 A current. If the d=18 cm and l is adjusted so that the magnetic field at C is zero, the value of the current l is A. 10.0 AB. 30.0AC. 8.0 AD. 18.0 A |
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Answer» Correct Answer - b The magnetic field at C due to first conductors is `B_(1)=(mu_(0))/(2pi) l/(3d//2)` (since point C is separated by `d+d/2=(3d)/2` from 1st conductors). The direction of field is perpendicular to the plane of paper and directed outwards. The magnetic field at C due to second conductors is `B_(2)=(mu_(0))/(2pi) 10/(d//2)`(since, point C is separated by `d/2` from 2nd conductors) the direction of field is perpendicular to the plane of paper and directed inwards. since direction of `B_(1)` and `B_(2)` at point C is in opposite direction and the magnetic field at C is zero , therefore, `B_(1)=B_(2)` `(mu_(0))/(2pi) l/(3d//2) =(mu_(0))/(2pi) 10/(d//2)` on solving `I=30.0 A ` |
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| 92. |
A conductor in the form of a right angle `ABC` with `AB=3 cm` and `BC = 4 cm` carries a current of `10 A`.There is a uniform magnetic field of `5 T` perpendicular to the palne of the conductor. The force on the conductor will beA. 1.5 NB. 2.0 NC. 2.5 ND. 3.5 N |
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Answer» Correct Answer - C |
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| 93. |
A proton moving with a constant velocity passes through a region of space without any changing its velocity. If `E` and `B` represent the electric and magnetic fields, respectively. Then, this region of space may haveA. `E=0, B=0`B. `E=0, B!=0`C. `E!=0, B=0`D. `E!=0, B!=0` |
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Answer» Correct Answer - (a,b,d) There is no change in velocity. It can be possible when electric and magnetic field are absent, i.e., `E=0, B=0`. Or when electric and magnetic fields are present but force due to electric field is equal and opposite to the due to magnetic field, (i.e.,`E!=0 B!=0`). Or when `E=0` but `B!=0`. `F=q upsilonB sin theta` i.e., `sin theta=0`, i.e., `theta=0 implies upsilon` and B are in the same direction. |
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| 94. |
Assertion: A proton an an alpha particle having the same kinetic energy are moving in circular paths in a unifrom magnetic field. The radii of their circular paths wll be equal. Reason: Any two charged particles having equal kinetic energies and entering a region of unifrom magnetic feild `vec(B)` in a direction perpendicular to `vec(B)`, will describe circular trajectoreis of equal radii.A. If both the assertion and reason are true and reason is a true explantion of the assertion.B. If both the assertion and reason are true but the reason is not the correct explantion of the assertion.C. If the assertion is true but reason falseD. If both the assertion and reason are false. |
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Answer» Correct Answer - A The ratio of the circular path is given by `r = (mv)/(qB) = (sqrt(2mK))/(qB)` , where `K = (1)/(2) mv^(2)` Since `K` and `B` are the same for the two particles, `r prop (sqrt(m))/(q)`, Now, the charge of an alpha particle is twice that of a proton and its mass is four times the mass of a proton, `sqrt(m)//q` will be the same for body particles. Hence, `r` will be the same for both particles. |
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| 95. |
Asertion: magnetic field due to a infinite stragith conductor varies inversely as the distance from it. Reason: the magentic field at the centre of the circular coil is zeroA. If both the assertion and reason are true and reason is a true explantion of the assertion.B. If both the assertion and reason are true but the reason is not the correct explantion of the assertion.C. If assertion is true but reason false.D. If asseration and reason both are false. |
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Answer» Correct Answer - B The magnetic field at a point due to current flowing thorugh an infinitely long conductor is given by `B = (mu_(0))/(4pi), (2l)/(a)` where, `a` is the distance of that from conductor. Now accroding to right hand thumb rule in follows that magnetic field is in the form of concentric circles, whose centres lie on the straight conductor (but opposite in direction) my symbol. |
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| 96. |
A current carrying circular loop is freely suspended by a long thread. The plane of the loop will point in the directionA. wherever left freeB. North-southC. East-southD. At `45^(@)` with the east-west direction |
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Answer» Correct Answer - C |
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| 97. |
A proton carrying `1 MeV` kinetic energy is moving in a circular path of radius `R` in unifrom magentic field. What should be the energy of an `alpha-` particle to describe a circle of the same radius in the same field?A. `2 MeV`B. `1 MeVC. `0.5 MeVD. `4 MeV |
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Answer» Correct Answer - B For proton `r = (sqrt(2m(KE)))/(qB)` So `q prop sqrt(m(KE))` Hence `(e)/(2e) = sqrt(((m_(p)) (1 meV))/((4m_(p)) (KE)))` `(1)/(4) = (1 MeV)/(4KE)` `KE = 1 MeV` |
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| 98. |
The circular current loop of radius b shown in Fig. is mounted rigidly on the axle, midway between the two supporting cords. In the absence of an external magnetic field, the tensions in the cords are equal and are `T_0.` a. What will be the tensions in the two cords when the vertical magnetic field B is present? b. Repeat if the field is parallel to the axis. |
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Answer» Let mass of hte hanging system is m, then `2T_0=mg to` in the absence of magnetic field (a) When magnetic field is applied: `tau_0=MB=pib^2IB....(i)` `T_1+T_2=mg...(ii)` Taking torque about O: `tau_0+T_2L/2=T_1L/2....(iii)` Solve equations (i), (ii) and (iii) to get `T_1=(mg)/2+(pib^2BI)/L and T_2=(mg)/2-(pib^2BI)/L` (b) In this case, magnetic field will produce no torque, so tension will remain same, `T_0=mg//2` |
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| 99. |
A current carrying loop is free to turn in a uniform magnetic field. The loop will then come into equilibrium vhen its plane is inclined atA. `0^(@)` to the direction of the fieldB. `45^(@)` to the direction of the fieldC. `90^(@)` to the direction of the fieldD. `135^(@)` to the direction of the field |
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Answer» Correct Answer - C |
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| 100. |
A proton `(mass = 1.67xx10^(-27) kg` and charge `1.6xx10^(-19) C)` enters perpendicular to a magentic field of intensity `2 weber//m^(2)` with a speed of `2.6xx10^(7) m//sec`. The acceleration of the proton should beA. `6.5xx10^(15) m//sec^(2)`B. `6.5xx10^(13) m//sec^(2)`C. `6.5xx10^(11) m//sec^(2)`D. `6.5xx10^(9) m//sec^(2)` |
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Answer» Correct Answer - A `F = ma = qvB` `implies a = (qvB)/(m) = (1.6xx10^(-19)xx2xx3.4xx10^(7))/(1.67xx10^(-27))` `= 6.5xx10^(15) m//sec^(2)` |
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