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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. |
In the figure shown there are two semicircles of radii `r_(1)` and `r_(2)` in which a current `i` is flowing. The magnetic induction at the centre `O` will be A. `(mu_(0)i)/(4)(r_(1)+r_(2))`B. `(mu_(0)i)/(4)(r_(1)-r_(2))`C. `(mu_(0)i)/(4)[(r_(1)+r_(2))/(r_(1)r_(2))]`D. `(mu_(0)i)/(4)[(r_(1)-r_(2))/(r_(1)r_(2))]` |
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Answer» Correct Answer - C |
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| 52. |
A steel wire of length l has a magnetic moment M. It is bent into a semicircular arc. What is the new magnetic moment?A. `mxxl`B. `M/l`C. `(2M)/(pi)`D. `M` |
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Answer» Correct Answer - a |
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| 53. |
The magnetic induction at centre O in the following figure will be A. `(mu_(0)ialpha)/(4pi) (1/(r_(1))-1/(r_(2))) o.`B. `(mu_(0)ialpha)/(4pi)(1/(r_(1))+1/(r_(2)))o.`C. `(mu_(0)ialpha)/(2pi)[1/(r_(1))-1/(r_(2))]ox`D. `(mu_(0)ialpha)/(2pi)[1/(r_(1))+1/(r_(2))]ox` |
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Answer» Correct Answer - A `B=(mu_(0))/(4pi).alpha.i[1/(r_(1))-1/(r_(2))]` |
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| 54. |
A current carrying wire is bent in the L shapes as shown in figure. The length of both arms extend to infinity. Then, the magnetic field at O is- A. zeroB. `mu_(0)I//2pir`C. `(mu_(0)I)/(2pir)(1+1/(sqrt(2)))`D. `(mu_(0)I)/(4pir)(1+1/(sqrt(2)))` |
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Answer» Correct Answer - C `B=2(mu_(0))/(4pir) [sin45^(@)+sin90^(@))=(mu_(0)I)/(2pir)(1+1/(sqrt(2)))` |
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| 55. |
An alectron in a circular orbit of radius 0.05 mn performs `10^(16) "rev"//s.` the magnetic moment due to this ratation of electron is `(in A-m^(2)).`A. `2.16xx10^(-23)`B. `3.21xx10^(-22)`C. `3.21xx10^(-24)`D. `1.26xx10^(-23)` |
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Answer» Correct Answer - d |
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| 56. |
How does the magnetic susceptibility of a paramagnetic material change with temperature?A. `chi prop T`B. `chi prop T^(-1)`C. `chi= "constant"`D. `chi prop e^(T)` |
| Answer» Correct Answer - B | |
| 57. |
The magnetic susceptibility of any paramagnetic material changes with absolute temperature `T` asA. `propT^(2)`B. `prop T^(1)`C. `prop T^(-1)`D. `prop T^(2)` |
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Answer» Correct Answer - C For paramagnetic materials, magnetic suspectibility is inversely proportional to the temperature i.e. proportional to `T^(-1)` |
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| 58. |
A wire carrying current `I` has the shape as shown in the adjoining figure. Linear parts of the wire are very long and parallel to X-axis while semicicular portion of radius `R` is lying in `Y-Z` plane. Magnetic field at point `O` is A. `B=(mu_(0))/(4pi)I/R(pihati+2hatk)`B. `B=(mu_(0))/(4pi)I/R(pihati-2hatk)`C. `B=(mu_(0))/(4pi)I/R(pihati+2hatk)`D. `B=(mu_(0))/(4pi)I/R(pihati-2hatk)` |
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Answer» Correct Answer - c |
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| 59. |
A strong magnetic field is applied on a stationary electron, thenA. moves in the direction of the fieldB. moves in an opposite direction of the fieldC. remains stationaryD. stats sprining |
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Answer» Correct Answer - C |
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| 60. |
A strong magnetic field is applied on a stationary electron, thenA. remains stationaryB. spins about its own axisC. moves in the direction of the fieldD. moves perpendicular to the direction of the field |
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Answer» Correct Answer - a |
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| 61. |
A particle of charge `q` and mass `m` moves in a circular orbit of radius `r` with angular speed `omega`. The ratio of the magnitude of its magnetic moment to that of its angular momentum depends onA. `omega` and `q`B. `omega`, `q` and `m`C. `q` and `m`D. `omega` and `m` |
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Answer» Correct Answer - C |
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| 62. |
A charge q moves region in a electric field E and the magnetic field B both exist, then the force on its isA. `qvec(v)xxvec(B)`B. `qvec(E)+qvec(v)xxvec(B)`C. `qvec(E)+qvec(B)xxvec(v)`D. `qvec(B)+q(vec(E)xxvec(v))` |
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Answer» Correct Answer - B If `vec(E)` is the electric field strength and `vec(B)` the magnetic field strength and q the charge on a particle, then electric force on the charge. `vec(F)_(e)=qvec(E)` and magnetic force on the charge `vec(F)_(m)=qvec(v)xxvec(B)` The net force on the charge `vec(F)=vec(F)_(e)+vec(F)_(m)=qvec(E)+qvec(v)xxvec(B)` |
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| 63. |
If a charged particle is a plane perpendicular to a uniform magnetic field with a time period T ThenA. `T^(2)propr^(3)`B. `T^(2)propr`C. `Tpropr^(2)`D. `Tpropr^(0)` |
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Answer» Correct Answer - D |
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| 64. |
Magnetic field due to a ring having n turns at a distance `x` on its axis is proportional to (if `r =` radius of ring)A. `(r)/((x^(2)+r^(2)))`B. `(r)/((x^(2)+r^(2))^(3//2))`C. `(nr^(2))/((x^(2)+r^(2))^(3//2))`D. `(n^(2)r^(2))/((x^(2)+r^(2))^(3//2))` |
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Answer» Correct Answer - C |
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| 65. |
A particle is projected in a plane perpendicular to a uniform magnetic field. The area bounded by the path described by the particle is proportional toA. the velocityB. the momentumC. the kinetic energyD. None of these |
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Answer» Correct Answer - C |
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| 66. |
If a electron velocity is (2hatixx3hatj)` and it subjected to a magnetic field `4hatk` thenA. speed of electon will chargeB. path of electron will chargeC. Both (a) and (b)D. None of above |
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Answer» Correct Answer - B |
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| 67. |
What is the net magnetic moment of an atom of a diamagnetic material?A. much greater than oneB. oneC. between zero and oneD. high retentivity and low coercive force |
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Answer» Correct Answer - D The magnetic momentum of a diamagnetic atom is equal to zero. |
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| 68. |
The desirable properties for making permanent magnets areA. high retenstivity, high coercivityB. hight retentivity, low coercivityC. low retentivity, high coercivityD. low retentivity, low coercivity |
| Answer» Correct Answer - A | |
| 69. |
The pole pieces of the magnet used in a pivoted coil galvanometer areA. plane surfaces of a bar magneticB. plane surfaces of a horse-shoe magnetC. cylindrical surfaces of a bar magnetD. cylindrical sufaces of a horse-shoe magnet |
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Answer» Correct Answer - D |
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| 70. |
Sometimes positive charged particle comes from space towards earth with high velocity. Its deviation due to the magnetic field of earth will be:A. towards northB. towards southC. towards westD. towards east |
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Answer» Correct Answer - D Positive charged particle are deviated towards east in earth magnetic field |
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| 71. |
If a positively charged particle is moving as shown in the figure, then it will be deflected due to magnetic field towards- A. `+x` directionB. `+y` directionC. `-x` directionD. `+z` direction |
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Answer» Correct Answer - D Magnetic force act along +z axis |
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| 72. |
A proton and an `alpha-`particle enter a uniform magnetic field moving with the same speed. If the proton takes `25 mu s` to make 5 revolutions, then the periodic time for the `alpha-` particle would beA. `50 mu sec `B. `25 mu sec`C. `10 mu sec`D. `5 mu sec` |
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Answer» Correct Answer - C `T=(2pim)/(qB)` `(T_(alpha))/(T_(P))=(m_(alpha))/(m_(p))xx(q_(p))/(q_(alpha))` `T_(alpha)=4/1xx1/2 xx5xx10^(-6)=10 mu sec` |
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| 73. |
Two parallel conductors `A` and `B` of equal lengths carry currents `I` and `10 I`, respectively, in the same direction. ThenA. A and B will repel each other with same forceB. A and B with attract each other with same forceC. A will attract B but B will repel AD. A and B will attract each other with different forces |
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Answer» Correct Answer - B |
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| 74. |
Two parallel beams of positrons moving in the same direction willA. not interact with each otherB. repel each otherC. attract each otherD. bedeflected normal to the plane containing two beams |
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Answer» Correct Answer - c |
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| 75. |
` H^(+), He^(+) and O^(++)` all having the same kinetic energy pass through a region in which there is a uniform magnetic field perpendicular to their velocity . The masses of ` H^(+), He^(+) and O^(2+)` are `1 amu, 4 amu and 16 amu` respectively . ThenA. `H^(+)` ions will be deflected mostB. `O^(2+)` ions will be deflected leastC. `He^(+)` and `O^(2+)` ions will suffer same deflectionD. All ions will suffer the same deflection |
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Answer» Correct Answer - A::C |
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| 76. |
Assertion A beam of electron can pass undeflected through a region of E and B. Reason Force on moving charged particle due to magnetic field may be zero in some cases. |
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Answer» Correct Answer - b |
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| 77. |
Assertion If the path of a charged particle in a region of uniform electric and magnetic field is not a circle, then its kinetic energy may remain constant. Reason In a combined electric and magnetic field region a moving charge experiences a net force `F=qE+q(VxxB),` where symbols have their usual meanings. |
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Answer» Correct Answer - b |
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| 78. |
If the angular momentum of an electron is `vec(J)` then the magnitude of the magnetic moment will beA. `(eJ)/m`B. `(eJ)/(2m)`C. eJ 2mD. `(2m)/(eJ)` |
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Answer» Correct Answer - B `M=pir^(2)i=(pir^(2)e)/T=(omegar^(2) e)/2=(omegamr^(2)e)/(2m)=(je)/(2m)` |
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| 79. |
If the velocity of charged particle has both perpendicular and parallel components while moving through a magnetic field ,then what is the path following by a charged particle?A. CircularB. EllipaticalC. LinearD. Helical |
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Answer» Correct Answer - d |
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| 80. |
A charged particle (charge `q`) is moving in a circle of radius `R` with unifrom speed `v`. The associated magnetic moment `mu` is given byA. `(qvR)/2`B. `qvR^(2)`C. `(qvR^(2))/2`D. qvR |
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Answer» Correct Answer - A As revolving charge is equivalent to a current, so, `i=qf=qxx(omega)/(2pi)` But `omega=V/R` Where R is radius of circle and v is uniform speed of charged particle. Therefore, `i=(qv)/(2piR)` Now, magnetic moment associated with charged particle is given by `mu=IA=IxxpiR^(2)` `mu=(qv)/(2piR)xxpiR^(2)=1/2 qvR` |
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| 81. |
Which of the follwing particles will have minimum frequency of revolution when projected with the same velocity perpendicular to a magnetic field?A. `Li^(+)`B. ElectronC. ProtonD. `He^(+)` |
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Answer» Correct Answer - A |
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| 82. |
A charged particle (charge `q`) is moving in a circle of radius `R` with unifrom speed `v`. The associated magnetic moment `mu` is given byA. `1/2 v^(2) R`B. `1/4 qvR`C. `1/2 aqR`D. `1/2 q^(2)vR` |
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Answer» Correct Answer - c |
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| 83. |
A charged particle of mass m and charge q describes circular motion of radius r in a unifrom magnetic fiels of strenght b the frequency of revolution isA. `(Bq)/(2pim)`B. `(Bq)/(2pim)`C. `(2pim)/(Bq)`D. `(Bq)/(2piq)` |
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Answer» Correct Answer - A |
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| 84. |
For protecting a sensitive equipment from the external magnetic field, it should be:A. placed inside an aluminium canB. placed inside an iron canC. wrapped with insulation around it when passing current through itD. surrounded with fine copper sheet |
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Answer» Correct Answer - B Iron is a ferromagnetic substance. There are no magnetic lines of force inside a frerromagnetic substance. So equipments may be protected by placing it inside the can be made of a ferromagnetic subtances. Hence it is placed inside the iron can. |
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| 85. |
Consider the following statements for a paramagnetic substance kept in a magnetic field (a) if the magnetic field increases, the magnetisation increases (b) magnetisation increasesA. both (a) and (b) are trueB. (a) is true but (b) is falseC. (b) is true but (a) is falseD. Both (a) and (b) are false |
| Answer» Correct Answer - B | |
| 86. |
Which of the following relations is not correct?A. `B=mu_(0) (H+1)`B. `B=mu_(0)H(1+chi_(m))`C. `mu_(0)=mu(1+chi_(m))`D. `mu_(r)=1+chi_(m)` |
| Answer» Correct Answer - C | |
| 87. |
Select the incorrect alternatives(s):A. has a fixed valueB. may be zeroC. may be infiniteD. may be negative |
| Answer» Correct Answer - A | |
| 88. |
The hysteresis cylcle for the material of a permanent magnet isA. short and wideB. tall and narrowC. fall and wideD. short and narrow |
| Answer» Correct Answer - A | |
| 89. |
In hydrogen atom, an electron is revolving in the orbit of radius `0.53 Å` with `6.6xx10^(15) rotations//second`. Magnetic field produced at the centre of the orbit isA. `0.125 Wb//m^(2)`B. `1.25 Wb//m^(2)`C. `12.5 Wb//m^(2)`D. `125 Wb/m^(2)` |
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Answer» Correct Answer - C |
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| 90. |
A particle of mass `1.6xx10^(-27) kg` and charge `1.6 xx 10^(-19)` coulomb enters a uniform magnetic field of 1 Tesla as shown in the figure. The speed of the particle is `10^7 m//s`. The distance PQ will be A. 0.14 mB. 0.28 mC. 0.4 mD. 0.5 m |
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Answer» Correct Answer - A `PQ=2r sin theta=2(mv)/(qB) sin theta=(2xx1.6xx10^(-27)xxsin45)/(1.6xx10^(-19)xx1)=2/(10sqrt(2)) =(sqrt(2))/10=0.14` |
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| 91. |
A particle of mass `m` and charge `q` moves with a constant velocity `v` along the positive `x` direction. It enters a region containing a uniform magnetic field `B` directed along the negative `z` direction, extending from `x = a` to `x = b`. The minimum value of `v` required so that the particle can just enter the region `x gt b` isA. `qpB//m`B. `q(b - a)B//m`C. `qaB//m`D. `q(b + a)B//2m` |
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Answer» Correct Answer - B |
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| 92. |
A proton moves at a speed `v = 2xx10^(6) m//s` in a region of constant magnetic field of magnitude B = 0.05 T. The direction of the proton when it enters this field is `theta = 30^(@)` to the field. When you look along the direction of the magnetic field, then the path is a circle projected on a plane perpendicular to the magnetic field. How far will the proton move along the direction of B when two projected circles have been completed?A. `4.35 m`B. `0.209 m`C. `2.82 m`D. `2.41 m` |
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Answer» Correct Answer - A |
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| 93. |
Two charged particles traverse identical helical paths in a completely opposite sense in a uniform magnetic field `vec(B)=B_(0)hat(K)`A. They have equal z-components of momentB. They must have equal chargesC. They necessarily represent a particles, anti-particle pairD. The charge to mass ratio satisfy `((e)/(m))_(1) + ((e)/(m))_(2) = 0` |
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Answer» Correct Answer - D |
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| 94. |
A charged particle enters in a uniform magnetic field perpendicular to it. Now match the following two columns. |
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Answer» Correct Answer - `A to p; B to q; C to q; D to q,s` |
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| 95. |
An electric current I enters and leaves a uniform circular wire of radius a through diametrically opposite points. A charged paricle q moving along the axis of the circular wire passes through its centre at speed v. The magnetic force acting on the particle when it passes through the centre has a magnitudeA. `qv(mu_(0)i)/(2a)`B. `qv(mu_(0)i)/(2pia)`C. `qv(mu_(0)i)/(a)`D. zero |
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Answer» Correct Answer - D |
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| 96. |
A charged particle enters a uniform magnetic field with velocity vector at angle of `45^@` with the magnetic field. The pitch of the helical path followed by the particle is p. the radius of the helix will beA. `p/(sqrt(2)pi)`B. `sqrt(2)p`C. `p/(2pi)`D. `(sqrt(2)p)/(pi)` |
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Answer» Correct Answer - C `P=(2pimv cos theta)/(qB)=(2pimv)/(sqrt(2)qB)` `R=(mv sin theta)/(qB)=(mv)/(sqrt(2)qB)=p/(2pi)` |
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| 97. |
A charged particle enters into a uniform magnetic field with velocity `v_(0)` perpendicular to it , the length of magnetic field is `x=sqrt(3)/(2)R`, where `R` is the radius of the circular path of the particle in the field .The magnitude of charge in velocity of the particle when it comes out of the field is A. `2v_(0)`B. `(v_(0))/(2)`C. `(sqrt(3)v_(0))/(2)`D. `v_(0)` |
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Answer» Correct Answer - D |
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| 98. |
A proton is moving in a uniform magnetic field B in a circular path of radius a in a direction perpendicular to Z- axis along which field B exists. Calculate the angular momentum. If the radius is a and charge on proton is e.A. `(Be)/(a^(2))`B. `eB^(2)a`C. `a^(2) eB`D. `aeB` |
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Answer» Correct Answer - c |
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| 99. |
An electron is moving on a circular path of radius `r` with speed `v` in a transverse magnetic field B. e/m for it will beA. `v/Br`B. `B/rv`C. BrvD. `vr/B` |
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Answer» Correct Answer - A |
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| 100. |
A particle of mass m, charge q and kinetic energy T enters in a transverse uniform magnetic field of induction B. After the 3 s, the kinetic energy of the particle will beA. 3TB. 2TC. TD. 4T |
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Answer» Correct Answer - C After pasing through a magnetic field, the magnitude of its mass and velocity of the particle remain same, so its energy does not change, ie, kinetic energy will remain T. |
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