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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 conductor PQ, carrying a current i is placed perpendicular to a long conductor xy carrying a current i. The direction of force on PQ will be A. Towards rightB. Towards leftC. UpwardsD. Downwards |
| Answer» Correct Answer - D | |
| 52. |
A conducting circular loop of radius r carries a constant current i. It is placed in a uniform magnetic field B such that B is perpendicular to the plane of loop. What is the magnetic force acting on the loop?A. `B I R`B. `2pi (B I R )`C. zeroD. `pi(BIR)` |
| Answer» Correct Answer - 3 | |
| 53. |
A conducting circular loop of radiius `r` carries a constant current `i`. It is placed in a uniform magnetic field `vec(B)_(0)` such that `vec(B)_(0)` is perpendicular to the plane of the loop . The magnetic force acting on the loop isA. `irB_(o)`B. `2piriB_(o)`C. zeroD. `piriB_(o)` |
| Answer» Correct Answer - 1 | |
| 54. |
A conducting circular loop of radiius `r` carries a constant current `i`. It is placed in a uniform magnetic field `vec(B)_(0)` such that `vec(B)_(0)` is perpendicular to the plane of the loop . The magnetic force acting on the loop isA. `ir B_(0)`B. `2pi irB_(0)`C. zeroD. `pi irB_(0)` |
| Answer» Correct Answer - C | |
| 55. |
In a region of crossed fields as shown, the strength of electric field is E and that of magnetic field is B. Three positively charged particles with speeds `V_(1), V_(2) and V_(3)` are projected and their paths are shown. From this it implies that A. `V_(1) gt (E)/(B)`B. `V_(2) = (E)/(B)`C. `V_(3) lt (E)/((B)`D. All of these |
| Answer» Correct Answer - A::B::C::D | |
| 56. |
In a region of space, both electric and magnetic field are present simultaneously in opposite direction. A positively charged particle is projected with certain speed an angle `theta (lt 90^(@))` with magnetic field. It will move in aA. Helicat path of uniform pitchB. Helical path of increasing pitchC. Helical path of decreasing pitchD. Helical paht, whose pitch first decreases and then increases |
| Answer» Correct Answer - D | |
| 57. |
The magnitude of the magnetic field at the centre of the tightly wound 150 turn coil of radius 12 cm carrying a current of 2 A isA. 18GB. 19.7GC. 15.7GD. 17.7G |
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Answer» Correct Answer - C Here, `N=150, R = 12 cm = 12xx10^(-2)m, I=2A` `therefore" "B=(mu_(0)NI)/(2R)=(2pixx10^(-7)xx150xx2)/(12xx10^(-2))=1.57xx10^(-3)T` `=1.57xx10^(-3)T=15.7xx10^(-4)T=15.7G` |
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| 58. |
A coil having N turns is wound tightly in the form of a spiral with inner and outer radii a and b respectively. A current I passes through the coil. The magnetic moment of the spriral isA. `M=(mu_(0)NI)/(3)(b^(3)-a^(3))/((b-a)^(3))`B. `M=(piNI)/(3)((b-a)^(3))/(b^(3)-a^(3))`C. `M=(piNI)/(3)((b^(3)-a^(3)))/((b-a)^(3))`D. `M=(mu_(0)NI)/(3)((b-a)^(3))/(b^(3)-a^(3))` |
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Answer» Correct Answer - C Magnetic moment for one loop of radius r is `M=IA=I(pir^(2))` Total number of turns per unit length is `(N)/(b-a)` `dM=(dN)(IA)dM=((N)/(b-a))dr` `(pi r^(2)I)=(piNI)/(b-a)r^(2)dr` `M=(piNI)/(b-a)int_(a)^(b)r^(2)dr=(piNI)/(b-a)(r^(3)/(3))_(a)^(b)=(piNI)/(3)((b^(3)-a^(3))/(b-a))` |
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| 59. |
A beam of protons is moving horizontally towards you. As it approaches, it passes through a magnetic field directed downward. The beam deflects- A. To your left sideB. To your right sideC. Does not deflectD. Nothing can be said |
| Answer» Correct Answer - B | |
| 60. |
A charged particle enters a region which offers a resistance against its motion and a uniform mangetic field exists in the region. The particle traces a spiral path as shown in Fig. 1.29. State the following statements as True or False. a. Component of magnetic field in the plane of spiral is zero. (b) Particle enters the region at Q. If magnetic field is outwards, then the particle is positively charged. (d) If magnetic field is outwards, then the particle is negatively charged. A. Component of magnetic field in the plane of spiral is zero.B. Particle enters the region at QC. If magnetic is outwards then the particle is positively charged.D. If magnetic field is outwards then the particle is negatively charged. |
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Answer» Correct Answer - A,B,C,D If particle has velocity component parallel to magnetic field path will be helical. Radius of path is decreasing so it enters at Q. |
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| 61. |
A beam of protons is deflected sideways. Could this deflection be caused (1) By an electric field ? (2) By a magnetic field ? (3) If either could be responsible how would you be able to tell which was present ? |
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Answer» (1) yes (2) yes (3) On reversing the direction of the projection of protons, if the protons are deflected in the same direction, then the deflection is due to electric field, if the protons are deflected in the opposite direction, then the deflection is due to magnetic field |
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| 62. |
There is region of space where uniform magnetic field of induction B exists. The field exists at all points for which `x-` coordingates are positive. The direction of field ia along negative z axis. Now a certain charged particle of mass m and charge q a certain speed enters in this region. A magnetic field at a point whose coordinates are `x=0,y=-d` and `z=0` . Magnetic force will start acting on the particle and particle moves in a uniform circular motion, such that origin becomes centre of circular path described by it The speed of the charge particle will beA. particle will come out from the region of uniform magnetic field at a point `(a=0,y=d,z=0)` with speed `(2qBd)/(m)`B. particle does not remain always in the `x-y` planeC. The `x-`coordinate of location of particle will be d at a time `t=(pi m)/(qB)`D. particle will emerge out of the region of field after a time `(pi m)/(qB)` with speed `(qBd)/(m)` |
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Answer» Correct Answer - D `theta=omegat, omega=(qB)/(m),pi=(qB)/(m).timpliest=(pim)/(qB)` |
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| 63. |
If a moving electron if deflected sideways in passing through a certain region of space, can we be sure that a magnetic field exists in that region? |
| Answer» No, the sideways deflection may be due to Electric field as well. In the absence of electric field, the sideways deflection shows the presence of magnetic field in the region | |
| 64. |
Figure shows a long straight conductor carrying current i. A wooden disc of mass m and radius R is placed near it such that conductor lies in the plane of the disc. The distance between the wire and centre of disc is `r (r gt R)`. A wire of length R and carrying current `i_(0)` is posted on the disc such that the wire is perpendicular to the long straight conductor. The disc is hinged at the centre. Neglecting mass of conducting wire, find (a) Angular acceleration of the disc at the instant (b) Reaction at the hinge at the instant |
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Answer» (a) `(mu_(0)ii_(0))/(pi mR^(2)) (R -r ln (r + R)/(r))` (b) `(mu_(0)ii_(0))/(2pi) ln ((r + R)/(r))` |
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| 65. |
Consider a system of two long coaxial cylinders: solid (of radius a carrying i current in positive z direction) and hollow (or radius b carrying current i in negative z direction). Find the magnetic field at a point distant r from of the cylinders for (i) `r le a` (ii) `a lt r lt b` (iii) `r gt b` |
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Answer» (i) For `r le a`, magnetic field in only due to solid cylinder. Thus it is, `B = (mu_(0)ir)/(2pi a^(2))` (ii) For `a lt r lt b`, magnetic field is only due to solid cylinder Thus, `B = (mu_(0)i)/(2pir)` (iii) For `r ge b`, magnetic field is due to both solid and hollow cylinders Thus, `B_(1) = (mu_(0)i)/(2pir)`, due to solid cylinder `B_(2) = (mu_(0)i)/(2pir)`, due to hollow cylinder Also `B_(1) and B_(2)` are in opposite directions `rArr "Net" B = B_(1) - B_(2) = 0` |
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| 66. |
Assertion : The net magnetic force on a current loop in a uniform magnetic field is zero but torque may or may not be zero. Reason : Torque on a current carrying coil in a magnetic field is given by `vec(tau)=nI(vec(A)Xxtau(B))`.A. If both assertion and reason are true and reason is the correct explanation of assertion.B. If both assertion and reason are true but reason is the correct explanation of assertion.C. If assertion is true but reason is falseD. If both assertion and reason are false. |
| Answer» Correct Answer - B | |
| 67. |
Find magnetic moment in following cases (Eash segment is of length l) |
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Answer» (i) `pil^(2)i` out of the page (ii) `(pir^(2))/(2) (hat(i) + hat(j))` (iii) `il^(2) (hat(i) + hat(j) + hat(k))` |
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| 68. |
An alpha particle moving with a velocity `vec(v)_(1)=2hat(i)m//s` is a uniform magnetic field experiences a force `vec(F)_(1)=-4e(hat(j)-hat(k))N`. When the particle moves with a velocity `vec(v)_(2)=hat(j)m//s`, then the force experienced by it is `vec(F)_(2)=2e(hat(i)-hat(k))N`. The magnetic induction `vec(B)` at that point is `:`A. `hat(i)+hat(j)-hat(k)`B. `-hat(i)-hat(j)+hat(k)`C. `hat(i)-hat(j)-hat(k)`D. `hat(i)+hat(j)+hat(k)` |
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Answer» Correct Answer - D `vec(B)=xhat(i)+yhat(j)+zhat(k)` `vec(F_(1))=2[2hat(i)xx(xhat(i)+yhat(j)+zhat(k))]=-4e(hat(j)-hat(k))` `vec(F_(2))=2e[2hat(j)xx(xhat(i)+yhat(j)+zhat(k))]=2e(hat(i)-hat(k))` `impliesvec(B)=hat(i)+hat(J)+hat(k)` |
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| 69. |
A current loop placed in a magnetic field behaves like aA. magnetic dipoleB. magnetic substanceC. magnetic poleD. non magnetic substance |
| Answer» Correct Answer - 1 | |
| 70. |
Find instantaneous direction of force in following cases (magnetic field and velocity are indicated ) |
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Answer» (i) Towards right (ii) Into the plane (iii) Upwards (iv) Towards left |
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| 71. |
A charge particle having charge 2 coulomb is thrown with velocity `2hat(i) + 3 hat(j)` inside a region having `vec(E) = 2 hat(j)` and magnetic field `5 hat(k)`. Find the initial Lorentz force acting on the particle |
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Answer» Lorentz force `vec(F) = q (vec(E) + vec(v) xx vec(B))` `vec(F) = 2 [2 hat(j) + (2 hat(j) + 3 hat(j)) xx 5 hat(k)]` `=2 [2 hat(j) + 10 (-hat(j)) + 15 hat(j)]` `= 30 hat(i) - 16 hat(j)` |
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| 72. |
Singly ionized heliium `(x)`, ionized deuteron`(y)`, alpha`(z)` particles are projected into a uniform magnetic field `3xx10^(-4)` tesla with velocities `10^(5)ms^(-1),0.4xx10^(4)ms^(-1)` and `2xx10^(3)ms^(-2)` respectively. The correct relation between the ration of the angular momentum to the magnetic moment of the particles isA. `xgty=z`B. `xltyltz`C. `yltxltz`D. `z gt x gt y` |
| Answer» Correct Answer - 1 | |
| 73. |
Find the instantaneous direction of force in following case (magnetic field and velocity are indicated) |
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Answer» As charge is positive, the direction of force is same as direction of `(vec(v) xx vec(B))`. So by Right hand thumb rule, curling the fingers of right hand from `vec(v) " to " vec(B)` thumb extends in direction of `-x`. Thus magnetic force at the instant shown is along `-hat(i)` direction. We can also do the problem by writing v and B in vector form. `vec(v) = v vec(j)` `vec(B) = B (-hat(k))` `vec(F) = q (vec(v) xx vec(B)) = q (v hat(j) xx B (-hat(k))` `= -qvB [hat(j) xx hat(k)]` `= - q v B hat(i)` Thus magnetic force is in `(-hat(i))` direction |
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| 74. |
A charged particle of charge 1 mC and mass 2g is moving with a speed of 5m/s in a uniform magnetic field of 0.5 tesla. Find the maximum acceleration of the charged particle |
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Answer» `m = 2 xx 10^(-3) kg, v = 5 m//s, B = 0.5` testa, `q = 1 xx 10^(-3) C` `vec(F) = q(vec(v) xx vec(B))` `rArr F = qvB sin theta` `rArr F = 10^(-3) xx 5 xx 0.5 xx sin 90^(@) N` `rArr F = 2.5 xx 10^(-3) N` `a = (F)/(m)` `rArr a = (2.5 xx 10^(-3))/(2 xx 10^(-3)) N//kg` `rArr a = 1.25 m//s^(2)` |
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| 75. |
A charged particle of charge `10mC` enters a uniform magnetic field of induction `bar(B)=4hat(i)+yhat(j)+zhat(k)` tesla with a velocity `bar(V)=2hat(i)+3hat(j)-6hat(k)`. If the particle continues to move undeviated then the strength of the magnetic field induction in teslaA. 4B. 8C. 14D. 30 |
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Answer» Correct Answer - 3 `vec(v)//vec(B)y=6z=-12|vec(B)|=14` |
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| 76. |
A charged particle moving in a magnetic field `vec(B) = hat(i) - hat(j)` tesla, has an acceleration of `2 hat(i) + alpha hat(j)` at some instant. Find the value of `alpha` |
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Answer» `vec(F) = q (vec(v) xx vec(B))` `vec(a) = (q)/(m) (vec(v) xx vec(B))` `rArr vec(a) bot vec(B)` Thus `vec(a).vec(B) = 0` `rArr (2 hat(i) + alpha hat(j)).(hat(i) - hat(j)) = 0` `rArr 2 - alpha = 0` `rArr alpha = 2` |
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| 77. |
Two long parallel wires are at a distance ` 2d` apart. They carry steady equal currents flowing out of the plane of the paper , as shown. The variation of the magnetic field `B` along the line `XX` is given byA. B. C. D. |
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Answer» Correct Answer - A The magnetic field varies inversely with the distance for a long conductor . That is, ` B prop (1)/(d)`. According to the magnitude and direction shown graph (1) is the correct one. |
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| 78. |
A current loop `ABCD` is held fixed on the plane of the paper as shown in figure. The arcs `BC( radius = b) and DA ( radius = a)` of the loop are joined by two straight wires `AB and CD` at the origin `O` is 30^(@)`. Another straight thin wire with steady current `I_(1)` flowing out of the plane of the paper is kept at the origin . The magnitude of the magnetic field (B) due to the loop `ABCD` at the origin (o) is :A. `(mu_(0)I ( b-a))/( 24 ab)`B. `(mu_(0)I)/( 4 pi) [(b-a)/(ab)]`C. `(mu_(0)I)/( 4 pi) [2(b-a) + pi//3( a + b)]`D. zero |
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Answer» Correct Answer - A The magnetic field at `O` due to current in `DA` is `B_(1) = ( mu_(0))/( 4 pi) (I)/(a) xx ( pi)/( 6)` ( directed vertically upwards) The magnetic field at `O` due to current in `BC` is `B_(2) = ( mu_(0))/( 4 pi) (I)/(b) xx ( pi)/( 6)` ( directed vertically downwards) The magnetic field due to current `AB and CD` at `O` is zero . Therefore the net magnetic field is `B = B_(1) - B_(2)` ( directed vertically upwards) ( mu_(0))/( 4 pi) (I)/(a) xx ( pi)/( 6) - ( mu_(0))/( 4 pi) (I)/(b) xx ( pi)/( 6) = (mu_(0)I)/( 24) ((1)/(a) - (1)/(b)) = (mu_(0)I)/( 24ab) (b-a)` |
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| 79. |
A current loop `ABCD` is held fixed on the plane of the paper as shown in figure. The arcs `BC( radius = b) and DA ( radius = a)` of the loop are joined by two straight wires `AB and CD` at the origin `O` is 30^(@)`. Another straight thin wire with steady current `I_(1)` flowing out of the plane of the paper is kept at the origin . Due to the process of the current `I_(1)` at the origin:A. The forcwes on `AD and BC` are zero.B. The magnitude of the net force on the loop is given by `(I_(1)I)/( 4 pi) mu_(0)[2 ( b-a)+pi//3( a+b)]`C. The magnitude of the net force on the loop is given by `(mu_(0)II)/( 24 ab) ( b-a)`D. The forces on `AB and DC` are Zero. |
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Answer» Correct Answer - A KEY CONCEPT : `vec(F) = I(vec(l)xxvec(B)) ` The force on `AD and BC` due to corrent `I_(1)` is zero. This is because the directions of current element `Ivec(dl)` and magnetic field `vec(B)` are parallel. |
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| 80. |
Relative permitivity and permeability of a material `epsilon_(r) and mu_(r)`, respectively . Which of the following values of these quantities are allowed for a diamagnetic material?A. `epsilon_(r ) = 0.5, mu_(r ) = 1.5`B. `epsilon_(r ) = 1.5, mu_(r ) = 0.5`C. `epsilon_(r ) = 0.5, mu_(r ) = 0.5`D. `epsilon_(r ) = 1.5, mu_(r ) = 1.5` |
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Answer» Correct Answer - B For a diamagnetic material , the value of `mu_( r )` is less than one. For any material , the value of `epsilon_(r)` is always greater than `1`. |
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| 81. |
A current loop `ABCD` is held fixed on the plane of the paper as shown in figure. The arcs `BC( radius = b) and DA ( radius = a)` of the loop are joined by two straight wires `AB and CD` at the origin `O` is 30^(@)`. Another straight thin wire with steady current `I_(1)` flowing out of the plane of the paper is kept at the origin . The magnitude of the magnetic field (B) due to the loop `ABCD` at the origin (o) is :A. zeroB. `(mu_(0)(b-a)i)/(24ab)`C. `(mu_(0)I)/(4pi)[(b-a)/(ab)]`D. `(mu_(0)I)/(4pi)[2(b-a)+(pi)/(3)[a+b)]` |
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Answer» Correct Answer - C Net magntic field due to loop `ABCD` at `O` is `B=B_(AB)+B_(BC)+B_(CD)+B_(DA)` `B=0+(mu_(0)I)/(4pia)xx(pi)/(6)+0-(mu_(0)I)/(4pib)xx(pi)/(6)` |
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| 82. |
A charged paricle goes undeflected in a region containing electric and magnetic field. It is possible thatA. ` 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 fields 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 magnetic force, `( i.e. E!= 0, B!= 0)` . * Or when ` E = 0. B!= 0 ` is provided ` F = qvB sin theta = 0` ` sin theta = 0 , i.e, theta = 0 rArr v and B` are in the same direction . |
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| 83. |
A charged particle goes undeflected in a region containing electric and magnetic field. It is possible that (i) `vec(E) || vec(B). vec(v) || vec(E)` (ii) `vec(E)` is not parallel to `vec(B)` (iii) `vec(v) || vec(B)` but `vec(E)` is not parallel to `vec(B)` (iv) `vec(E)||vec(B)` but `vec(v)` is not parallel to `vec(E)` |
| Answer» `vec(E) || vec(B), vec(v)|| vec(E) or vec(E)` is not parallel to `vec(B)` | |
| 84. |
A charged particle goes undeflected in a region containing electric and magnetic field. It is possible that (i) `vec(E) || vec(B). vec(v) || vec(E)` (ii) `vec(E)` is not parallel to `vec(B)` (iii) `vec(v) || vec(B)` but `vec(E)` is not parallel to `vec(B)` (iv) `vec(E)||vec(B)` but `vec(v)` is not parallel to `vec(E)`A. `vec(E)||vec(B),vec(v)||vec(E)`B. `vec(E)` is not parallel to `vec(B)`C. `vec(v)||vec(B)` but `vec(E)` is not parallel to `vec(E)`D. `vec(E)||vec(B)` but `vec(v)` is not parallel to `vec(E)`. |
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Answer» Correct Answer - A,B,C,D If `B & E & v` are in same direction then particle will be undeflected cause `vec(F_(M))=0` `(b)` is also possible is not parallel to `vec(B)` If `vec(F_(M))=qvB` and `vec(F_(E))=qE` then v is `_|_` to `vec(E)` and v is `_|_` to `vec(B) vec(E)` is not parallel to `vec(B)` |
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| 85. |
A charged paricle goes undeflected in a region containing electric and magnetic field. It is possible thatA. `vec(E) || vec(B), vec(V)|| vec(E)`B. `vec(E)` is not parallel to `vec(B)`C. `vec(V) || vec(B), vec(E)` is not parallel to BD. `vec(E)|| vec(B)`, but V is not parallel to `vec(B)` |
| Answer» Correct Answer - A::B | |
| 86. |
A coil of area `100cm^(2)` having 500 turns carries a current of `1mA` . It is suspended in a uniform magnetic field of induction `10^(-3)wb//m^(2)` . Its plane makes an angle fo `60^(@)` with the lines of induction. The torque acting on the coil isA. `250xx10^(-8)Nm`B. `25xx10^(-8)Nm`C. `2.5xx10^(-8)Nm`D. `0.2xx10^(-8)Nm` |
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Answer» Correct Answer - 1 `tau=BAIN cos theta` |
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| 87. |
A square coil of side `10cm` consists of 20 turns and carries a current of 12A. The coil is suspended vertically and normal to the plane of the coil makes an angle of `30^@` with the direction of a uniform horizontal magnetic field of magnitude `0*80T`. What is the magnitude of torque experienced by the coil?A. 1.6 N mB. 1.2 N mC. 1.4 N mD. 1.8 N m |
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Answer» Correct Answer - A `tau=NIABsin theta` Here, `N=25, I=10A, B=0.9T, theta=30^(@)` `A=a^(2)=12xx10^(-2)xx12xx10^(-2)=144xx10^(-4)m^(2)` `therefore" "tau=25xx10xx144xx10^(-4)xx0.9xxsin 30^(@)="1.6 N m"` |
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| 88. |
A solenoid 50 cm long has 4 layers of windings of 350 turns each. If the current carried is 6A. Find the magnetic field at the centre of the solenoid |
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Answer» Here length `= 50 cm = 0.5 cm` `i = 6 A` Number of turns per unit length `= n = (4 xx 350)/(0.5) = 2800 m^(-1)` Magnetic field at the centre `= mu_(0) ni` `= 4 pi xx 10^(-7) xx 2800 xx 6` `= 2.11 xx 10^(-2) T` |
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| 89. |
Two long wires carrying current `I_1` and `I_2` are arranged as shown in figure. The one carrying current `I_1` is along the x-axis. The other carrying current `I_2` is along a line parallel to the y-axis given by `x=0` and `z=d`. Find the force exerted at `O_2` because of the wire along the x-axis. A. BilB. `2Bil`C. ZeroD. `sqrt(2)` Bil |
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Answer» Correct Answer - 3 According to right hand thumb rule, the magnetic field due to current `I_(1)` in the wire at point `O_(2)` is along `y-` axis. We know, magnetic force, `vec(F)_(m)=Ivec(l)xxvec(B)` Here, `vec(B)=-Bhat(y) and hat(l)=lhat(y)` `:. vec(F)_(m)=-IlB(hat(y)xxhat(y))=vec(0)` |
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| 90. |
The deflection in a moving coil galvanometer falls from 100 divisions to 20 divitation when a shunt of `12 Omega` is used. The resistance of the galvanometer coilds isA. `3Omega`B. `12Omega`C. `48 Omega`D. `48//5Omega` |
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Answer» Correct Answer - 3 `S=(G)/(n-1)` |
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| 91. |
A galvanometer of resistance `40 Omega` gives a deflection of 5 divisions per mA. There are 50 divisions on the scale. The maximum current that can pass through it when a shunt resistance of `2Omega` is connected isA. 210mAB. 155 MaC. 420mAD. 75 Ma |
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Answer» Correct Answer - A `I_(G) = 50/5=10 mA, R_(G)=40 Omega, R_(S)= 2 Omega` Maximum current, `I=(R_(G) + R_(S))/(R_(G)) xx I_(G) = ((40 + 2) xx 10)/(2) = 210 mA` |
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| 92. |
A galvanometer having a coil resistance of `100 omega` gives a full scale deflection , when a current of `1 mA` is passed through it. The value of the resistance, which can convert this galvanometer into ammeter giving a full scale deflection for a current of `10A` , is :A. `0.1 omega`B. ` 3 omega`C. ` 0.01 omega`D. ` 2 omega` |
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Answer» Correct Answer - C `Ig G = (I - Ig)s` :. `10^(-3) xx 100 = ( 10 - 10^(-3)) xx S` :. S ~~ 0.01 omega` |
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| 93. |
A galvanometer of internal resistance `100 Omega` has a full scale deflection current of `1mA`. To convert it into a voltmeter of range `0-10V`, the resistance to be connected isA. `9000 Omega` in SeriesB. `10,000 Omega` in SeriesC. `9,900 Omega` in SeriesD. `9,800 Omega` in Series |
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Answer» Correct Answer - 3 `R=(V)/(i_(g))-G` |
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| 94. |
A long straight wire carrying current of `30 A` is placed in an external unifrom magnetic field of induction `4xx10^(4) T`. The magnetic field is acting parallel to the direction of current. The maggnetic of the resultant magnetic inuduction in tesla at a point `2.0 cm` away form the wire isA. `10^(-4)`B. `3xx10^(-4)`C. `5xx10^(-4)`D. `6xx10^(-4)` |
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Answer» Correct Answer - 3 `B=sqrt(B_(1)^(2)+B_(2)^(2))` |
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| 95. |
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 - 2 Force on the wire `F=Bil =2xx2xx3xx10^(-2)` |
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| 96. |
A proton of energy `2 MeV` is moving perpendicular to uniform magnetic field of `2.5T`. The form on the proton is `(Mp=1.6xx10^(-27)Kg` and `q=e=1.6xx10^(-19)C)`A. `2.5X 10^(-10)` newtonB. `8X 10^(-11)` newtonC. `2.5X 10^(-11)` newtonD. `8X 10^(-12)` newton |
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Answer» Correct Answer - 4 `F=Bqv,v=sqrt((2KE)/(m))` |
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| 97. |
A toroidal solenoid has `3000` turns and a mean radius of `10cm`. It has a soft iron core of relative magnetic permeability `2000`. Find the magnitude of the magnetic field in the core when a current of `1.0amp.` is passed through the solenoid.A. `20T`B. `12T`C. `6T`D. `3T` |
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Answer» Correct Answer - B `B=(muNi)/(2pir)` |
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| 98. |
A solenoid of length 0.6 m has a radius of 2 cm and is made up of 600 turns If it carries a current of 4 A, then the magnitude of the magnetic field inside the solenoid isA. `6.024xx10^(-3)T`B. `8.024xx10^(-3)T`C. `5.024xx10^(-3)T`D. `7.024xx10^(-3)T` |
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Answer» Correct Answer - C Here, `n=(600)/(0.6)="1000 turns/m, I = 4 A"` `"l = 0.6 m, r = 0.02 m "because (l)/(r)=30` i.e., `l gt gt r` Hence, we can use long solenoid formula, then `therefore " "B=mu_(0)nI=4pi xx10^(-7)xx10^(3)xx4=50.24 xx10^(-4)` `=5.024 xx10^(-3)T` |
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| 99. |
A solenoid of length `0*5m` has a radius of `1cm` and is made up of `500` turns. It carries a current of `5A`. What is the magnitude of the magnetic field inside the solenoid?A. `3.14xx10^(-3)T`B. `6.28xx10^(-3)T`C. `9.14xx10^(-3)T`D. `1.68xx10^(-3)T` |
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Answer» Correct Answer - 2 `B=mu_(0)i n` |
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| 100. |
The conversion of a moving coil galvanometer into a voltmeter is done byA. introducing a resistance of large value in seriesB. introducing a resistance of small value in parallelC. introducing a resistance of large value in parallelD. introducing a resistance of small value in series |
| Answer» Correct Answer - A | |