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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.
| 151. |
In previous problem, if the current is I and the magnetic field at D has magnitude B, thenA. `B=(mu_0I)/(2sqrt2pi)`B. `B=(mu_0I)/(2sqrt3pi)`C. B is parallel to the z-axisD. B makes an angle of `45^@` with the x-y plane |
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Answer» Correct Answer - (a,d) The magnitude of the magnetic field depends only on the distance from the x-axis. Point A and C are at distance of 1 unit each from the x-axis. Points B and D are at distance of `sqrt2` unit each from the x-axis. Magnetic field at point D, `B=(mu_0I)/(1sqrt2pi)` It is obvious that B is inclined at an angle of `45^@` with the x-y plane. |
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| 152. |
A magnet of magnetic moment `20` C.G.S. units is freely suspended in a uniform magnetic field of intensity 0.3 C.G.S. units. The amount of work done in deflecting it by an angle of `30^(@)` in C.G.S. unit isA. 6B. `3sqrt(3)`C. `3(2-sqrt(3))`D. 3 |
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Answer» Correct Answer - c Work done , `W=MB_(mu)(2-cos theta)` `=20xx0.3(1-cos 30^(@))=6(1-(sqrt(3))/2)=3(2-sqrt(3))` |
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| 153. |
Two wires wrapped over a conical frame from the coils `1` and `2`. If they produce no net magnetic field at the apex `P`, the value of A. `sqrt((r_(2))/(r_(1)))`B. `(r_(2))/(r_(1))`C. `((r_(1))/(r_(2)))^(2)`D. `sqrt((r_(1))/(r_(2)))` |
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Answer» Correct Answer - B `B=(mu_(0))/(4pi) (2M)/(x^(3))=(mu_(0))/(4pi) (ipir^(2))/((r//sin theta)^(3))` `implies B prop i/r B_(1) =B_(2) implies (i_(1))/(r_(1))=(i_(2))/(r_(2)) implies (i_(1))/(i_(2))=(r_(1))/(r_(2))` |
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| 154. |
Two wires are wrapped over wooden cylinder to form two co-axial loops carrying currents `i_(1)` and `i_(2)`. If `i_(2)=8i_(1)` the value of `x` for `B=0` at the origin `O` is: A. `(sqrt(7)-1)R`B. `sqrt(5) R`C. `sqrt(3) R`D. `sqrt(7) R` |
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Answer» Correct Answer - D `B_(1)=B_(2) implies (mu_(0))/(4pi) (2i_(1)piR^(2))/((R^(2)+R^(2))^(3//2))=(mu_(0))/(4pi) (2i_(2)piR^(2))/((x^(2)+R^(2))^(3//2))` given `i_(2)=8i_(1)` Solve to get `x=sqrt(7)R` |
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| 155. |
A charged particle moves with velocity `vec v = a hat i + d hat j` in a magnetic field `vec B = A hat i + D hat j.` The force acting on the particle has magnitude F. Then,A. `F=0, if aD=dA`.B. `F=0, if aD=-dA`.C. `F=0, if aA=-dD`.D. `fprop (a^2+b^2)^(1//2)xx(A^2+D^2)^(1//2)` |
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Answer» Correct Answer - a `vecFprop(vecVxxvecB)=hatk[aD-dA]` |
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| 156. |
A charged particle has acceleration `vec a = 2 hat i + x hat j ` in a megnetic field `vec B = - 3 hat i + 2 hat j - 4 hat k.` Find the value of x. |
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Answer» `vecFbotvecB :. vecabotvecB:. veca.vecB=0` `:. (2hati+xhatj).(-3hati+2hatj-4hatk)=0 implies -6+2x=0 implies x=3`. |
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| 157. |
Two parallel beams of positrons moving in the same direction willA. attract each otherB. Repel each otherC. not interact with each otherD. be deflected normal to the plane containing the two beams |
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Answer» Correct Answer - A Positron is the antiparticle or the antimatter counterpart of the electron. The position has an electron. The positron has an electron charge `+e`. Since direction of current is in the direction of positroon, it is experimentally observed that the two current carrying conductors attract each other. |
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| 158. |
A long wire `AB` is placed on a table. Another wire `PQ` of mass `1.0 g` and length `50cm` is set to slide on two rails `PS` and `QR`. A current of `50 A` is passed through the wires. At what distance above `AB`, will the wire `PQ` be in equilibrium? A. `25mm`B. `50mm`C. `75mm`D. `100mm` |
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Answer» Correct Answer - A Suppose in equilibrium wire `PQ` lies at a distance `r` above the wire `AB` Hence in equilibrium `mg=Bilimpliesmg=(mu_(0))/(4pi)((2i)/(r )) xxil` ` implies10^(-3)xx10=10^(-7)xx(2xx(50^(2)))/(r )=0.5impliesr=25mm` |
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| 159. |
The magnetic field due to a conductor fo unifrom cross section of radius `a` and carrying a steady current is represented byA. B. C. D. |
| Answer» Correct Answer - A | |
| 160. |
The ratio of the energy required to set up in a cube of side 10 cm, a uniform field of 4 `Wb//m^(2)` and a uniform electric field of `10^(6) V//m` isA. `1.44xx10^(7)`B. `1.44xx10^(-5)c`.C. `1.44xx10^(6)`D. `1.44xx10^(3)` |
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Answer» Correct Answer - c `U_(E)=1/2 epsilon_(0)E^(2)xx `volume `U_(B)=(B^(2))/(2mu_(0)) xx` volume `:. (U_(B))/(U_(E))=(B/E)^(2) 1/(mu_(0)epsilon_(0))=(4/(10^(6)))^(2) (3xx10^(8))^(2)` `=16/(10^(12)) xx9xx10^(16) = 144xx10^(4) =1.44xx10^(6)` |
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| 161. |
An electric field of 1500 V/m and a magnetic field of 0.40 Wb/`m^(2)` act on a moving electron. The minimum uniform speed along a straight line, the electron could have isA. `1.6xx10^(15) m//s`B. `6xx10^(-16) m//s`C. `3.75xx10^(3) m//s`D. `3.75xx10^(2) m//s` |
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Answer» Correct Answer - c Here , `E=1500 V//m , B=0.4Wb//m^(2)` Minimum speed of electron along the straight line, `v=E/B` `=1500/0.4 =3750=3.75xx10^(3) m//s` |
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| 162. |
A uniform magnetic field and a uniform electric field are produced, pointing in the same direction. An electron is projected with its velocity pointed in the same direction. What will be the effect on electron?A. The electron will turn to its rightB. The electron will turn to its leftC. The electron velocity will increase in magnitudeD. The eletron velocity wil decrease in magnitude |
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Answer» Correct Answer - D |
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| 163. |
A particle of charge -q and mass m enters a uniform magnetic field `vecB` (perpendicular to paper inward) at P with a velocity `v_0` at an angle `alpha` and leaves the field at Q with velocity v at angle `beta` as shown in fig. A. `alpha`=`beta`B. `v=v_0`C. `PQ=(2mv_0sin alpha)/(Bq)`D. The particle remains in field for time `t=(2m(pi-alpha))/(Bq)` |
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Answer» Correct Answer - (a,b,c,d) Option (b) is obvious `r=(mv_0)/(qB)` `PQ=2r sin alpha=2(mv_0)/(qB) sin alpha` `alpha=beta` Time taken `=T=(2pir)/v` `T=(2pim)/(qB)` For t time, `t=T/(2pi)(2pi-2alpha)=(2m)/(qB)(pi-alpha)` |
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| 164. |
Velocity and acceleration vector of a charged particle moving in a magnetic field at some instant are `vecv=3hati+4hatj and veca=2hati+xhatj`. Select the correct options.A. `x=-1.5`B. `x=3`C. Magnetic field is along z-directionD. Kinetic energy of the particle is constant |
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Answer» Correct Answer - (a,b,c) As `vecFbot vecv, so vecF.vecv=0 implies mveca.vecv=0` `implies m(2hati+xhatj).(3hati+4hatj)=0` `implies 6+4x=0 implies x=-1.5` Let the magnetic field is `vecB=ahati+bhatj+chatk`, then from `vecF=qvecv xx vecB` `implies m(2hati+xhatj)=q(3hati+4hatj)xx(ahati+bhatj+chatk)` `implies 2mhati-1.5mhatj=q(3bhatk-3chatj-4ahatk+4chati)` `4c=2m implies c!=0` `3b-4a=0 implies b=4/3a` |
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| 165. |
A circular coil of radius `R` carries an electric current. The magnetic field due to the coil at a point on the axis of the coil located at a distance `r` from the centre of the coil, such that `r gtgt R`, varies asA. `(1)/(r)`B. `(1)/(r^(3))`C. `(1)/(r^(2))`D. `(1)/(r^(3//2))` |
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Answer» Correct Answer - B The magnetic field due to a cicular coil of radius `R` at a point on the axis of the coil located at a distance `r` from the centre of the coil, `B = (mu_(0))/(4pi) (2pi I R^(2))/((R^(2) + r^(2))^(3//2))` Given `r gtgt R` then we have, after neglecting `R`, `B = (mu_(0))/(4pi) (2pi i R^(2))/(r^(3))` Also area `= pi R^(2)` |
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| 166. |
A circular coil having mass m is kept above ground (x-z plane) at some height. The coil carries i in the direction shown in Fig. 1.143. In which direction a uniform magnetic field `vec B` be applied so that the magnetic force balances the weight of the coil? A. Positive x-directionB. Negative x-directionC. Positive z-directionD. None of these |
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Answer» Correct Answer - d Net force on a current carrying loop in a uniform magnetic field is zero. So, magnetic force cannot balance its weight. |
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| 167. |
A circular coil having mass m is kept above ground (x-z plane) at some height. The coil carries i in the direction shown in Fig. 1.143. In which direction a uniform magnetic field `vec B` be applied so that the magnetic force balances the weight of the coil? A. positive `x`-directionB. negative `x`-directionC. positive `z`-directionD. None of these |
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Answer» Correct Answer - D Net force on a current carrying loop in a uniform magnetic field is zero. So magnetic force can not balance its weight. |
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| 168. |
Two parallel wires in free spaces are `10cm` apart and each carries a current of `10A` in the same direction. The force one wire exerts on the other per metre of length isA. `2xx10^(-4)N` , attractiveB. `2xx10^(-4)N` ,repulsiveC. `2xx10^(-7)N` ,attractiveD. `2xx10^(-7)N` ,repulsive |
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Answer» Correct Answer - A `F=(mu_(0))/(4pi)(2i_(1)i_(2))/(a)=10^(-7)xx(2xx10xx10)/(0.1)=2xx10^(-4)N` Direction of current is same, so force is attractive. |
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| 169. |
A galvanometer of resistance `G` is shunted by a resistance `S ohm`. To keep the main current in the circuit uncharged, the resistnace to be put in series with the galvonmeterA. `(S^(2))/((S + G))`B. `(SG)/((S + G))`C. `(G^(2))/((S + G))`D. `(G)/((S + G))` |
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Answer» Correct Answer - C Current will be uncharged if resistance remain the same, so `G = (GS)/(G + S) + R` `R = G - (GS)/(G + S) implies R = (G^(2))/(G + S)` |
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| 170. |
A slightly divergent beam of charged particles accelerated by a Potential difference V propogates from a point A along the axis of a solenoid. The beam is brought into focus at a distance l from the point A at two successive values of magnetic induction `B_1 and B_2.` Find the specific charge `q//m` of the particles. |
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Answer» Let us first calculate the velocity of the particles from the energy equation, `1/2mv^2=Vq implies v=sqrt((2Vq)/m)` Since the charged particles are slightly divergent, they will follow a helical path. Let `theta` be the small angle made by a particle with `B cos theta~=1` `:.` P(pitch of the particle)=`v_(||)xxT` `=v cos thetaxx(2pim)/(qB)=(2pivm)/(qB)` Particles are focussed if l contains integral number of pitches. `l=np implies p=l//n= l, l//2, l//3,....` `:.` for two consecutive focussing (as B increases, p decrease) `l=(2pimv)/(qB_1) and l/2=(2pimv)/(qB_2)` or `B_1=(2pimv)/(ql) and B_2=(4pimv)/(ql)`, or `B_2-B_1=(2pimv)/(ql) or B_2-B_1=(2pim)/(ql)sqrt((2Vq)/m),` or `q/m=(8pi^2V)/(l^2(B_2-B_1)^2)` |
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| 171. |
Two particle X and Y having equal charge, after being accelerated through the same potential difference enter a region of uniform magnetic field and describe circular paths of radii `R_1 and R_2` respectively. The ratio of the mass of X to that of Y isA. `(R_1//R_2)^(1//2)`B. `R_2//R_1`C. `(R_1//R_2)^(2)`D. `R_1//R_2` |
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Answer» Correct Answer - c `R=(sqrt(2qVm))/(Bq) or Rprop sqrtm` `(R_1)/(R_2)=sqrt((m_X)/(m_Y)) or (m_X)/(m_Y)=((R_1)/(R_2))^2` |
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| 172. |
The charge on a particle `Y` is double the charge on particle `X`.These two particles `X` and `Y` after being accelerated through the same potential difference enter a region of uniform magnetic field and describe circular paths of radii `R_(1)` and `R_(2)` respectively. The ratio of the mass of `X` to that of `Y` isA. `((2R_(1))/(R_(2)))^(2)`B. `((R_(1))/(2R_(2)))^(2)`C. `(R_(1)^(2))/(2R_(2)^(2))`D. `(R_(1))/R_(2)` |
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Answer» Correct Answer - C |
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| 173. |
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. `vec(B) = (mu_(0))/(4pi) (I)/(R) (pi hat(i) + 2 hat(k))`B. `vec(B) = - (mu_(0))/(4pi) (I)/(R) (pi hat(i) - 2 hat(K))`C. `vec(B) = -(mu_(0))/(4pi) (I)/(R) (pi hat(i) + 2 hat(k))`D. `vec(B) = (mu_(0))/(4pi) (I)/(R) (pi hat(i) - 2 hat(K))` |
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Answer» Correct Answer - C Magnetic field due to staright wire 1 `vec(B)_(1) = (mu_(0) I)/(4pi R) [sin 90^(@) +sin 0^(@)] (-vec(k))` `=(-mu_(0)I)/(4pi R)(hatk)=vec(B)_(3)` Magnetic field due to semicircular wire 2 `B_(2)= (-mu_(0) I)/(4R) (-hat(i)) = (-mu_(0) I)/(4pi R) (pi hat(i))` Magnetic field `vec(B)` at centre, `vec(B)_(c) = vec(B)_(1) + vec(B)_(2) + vec(B)_(3)` `implies vec(B)_(c) = (-mu_(0) I)/(4pi R) (pi hat(i) + 2 hat(k))` |
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| 174. |
A very long straight wire carries a current I. At the instant when a charge `+Q` at point `P` has velocity `vecV`, as shown, the force on the charge is A. Opposite to OXB. Along OXC. Opposite to OYD. Along OY |
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Answer» Correct Answer - D |
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| 175. |
In Fig. , AB is a non-conducting rod. Equal charges of magnitude q are fixed at various points on the rod as shown. The rod is rotated uniformly about an axis passing through O and perpendicular to its length such that linear speed at the end A or B of the rod is `3 ms^(-1)`. Magnetic field at O is A. `(11mu_(0)q)/(12 pi)`B. `(3mu_(0)q)/(7 pi)`C. `(mu_(0)q)/(2 pi)`D. `(6mu_(0)q)/(13 pi)` |
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Answer» Correct Answer - A Since the rod is a rigid a rigid body, every point has same angular velocity `W=1 rad//s`. For points at a distance of `1m, v_(1)=1m//s` For points at a distance of `2m, v_(2)=2m//s` If a charge rotating in circle constitues a current `I`, `B=(mu_(0)i)/(2r)`, where `iq/t=(qv)/(2pir)` Since field due to rotation of all charges are in same direction `B=Sigma (mu_(0)qv)/(4pir^(2)) =(mu_(0)q)/(4pi) (1+1+1/2+1/2+1/3+1/3)` `implies B=(11 mu_(0)q)/(12pi)` |
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| 176. |
Consider a hypothetic spherical body. The body is cut into two parts about the diameter. One of hemispherical portion has mass distribution m while the other portion has indentical charge distribution q. The body is rotated about the axis with constant speed `omega.` Then, the ratio of magnetic moment to angular momentum is A. `q/(2m)`B. `gtq/(2m)`C. `lt q/(2m)`D. cannot be calculated |
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Answer» Correct Answer - a The ration `M//L` is always `(q)/(2m)`. |
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| 177. |
Three infinitely long wires, each carrying a current 1A, are placed such that one end of each wire is at the origin, and, one of these wires is along x-axis, the other along y-axis and the third along z-axis. Magnetic induction at point (-2 m, 0, 0) due to the system of these wires can be expressed as A. `(mu_(0))/(4pi)(hatj+hatk)`B. `(mu_(0))/(4pi)(hatj-hatk)`C. `(mu_(0))/(8pi)(-hatj+hatk)`D. `(mu_(0))/(8pi)(hatj+hatk)` |
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Answer» Correct Answer - C The magnetic field due to wire placed along `x`-axis will be zero. The point under consideration is semi finitite position of the wires placed along `y`-axis and `z`-axis. Magnetic field due to wire placed along `y`-axis `vec(B)_(y)=(mu_(0)I)/(4pir) (hatK)` Magnetic field due to wire placed along `z`-axis `vec(B)_(z)=(muI)/(4pir) (-hatj)` Hence net magnetic field `vec(B)=(mu_(0)I)/(2pir) (hatk-hatj)` After subtituting the values we get `vec(B)=(mu_(0))/(8pi) (hatk-hatj)` |
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| 178. |
A proton of mass m and charge `+e` is moving in a circular orbit in a magnetic field with energy `1 MeV`. What should be the energy of alpha-particle (mass=`4m` and charge=`+2e`), so that it can revolve in the path of same radius?A. 1 MeVB. 4 MeVC. 2 MeVD. 0.5 MeV |
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Answer» Correct Answer - A |
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| 179. |
An electron and a proton each travel with equal speeds around circular orbits in the same uniform magnetic field as indicated (not to scale) in Fig. The field is into the page on the diagram. The electron travels ____ arount the _____circle. A. clockwise, smaller,couterclockwise, largerB. couterclockwise, larger,clockwise,smallerC. clockwise,larger,couterclockwise,smallerD. couterclockwise,larger,clockwise,smaller |
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Answer» Correct Answer - a `r=(mv)/(qB) implies rpropm` (for v same for both sense) |
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| 180. |
An `alpha` particle and a proton travel with same velocity in a magnetic field perpendicular to the direction of their veloccites find the ratio of the radii of their circular pathA. `4:1`B. `1:4`C. `2:1`D. `1:2` |
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Answer» Correct Answer - C `r = (mv)/(q B) implies (r_(0))/(r_(p)) = (m_(alpha))/(q_(alpha)) = (4)/(1) xx (1)/(2) = (2)/(1)` |
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| 181. |
Circular loop of a wire and a long straight wire carry current `I_(c )` and `I_(e )`respectively as shown in figure. Assuming that these are placed in the same plane. The magnetic field will be zero at the centre of the loop when the separation `H` is: A. `(pi I_(c))/(I_(e) R)`B. `(I_(c) R)/(I_(e) pi)`C. `(mu_(0) r)/(4r)`D. `(I_(e) R)/(I_(e) pi)` |
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Answer» Correct Answer - D Magnetic field at the centre `O` of the loop of radius `R` is given by `B_(1) = (mu_(0) I_(e))/(2R)` where `I_(e)` is the current flowing in the loop. Magnetic fied due to stragiht current carrying wire at a distance `H, i.e.,` at the point `O` is given by `B_(2) = (mu_(0) I_(e))/(2pi H)` |
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| 182. |
Two parallel, long wires carry currents `t_(1) "and" t_(2)` with `t_(1)gtt_(2)`. When the currents are in the same direction, the magnetic field at a point midway between the wires is 10 `mu`T. If the direction of `i_(2)` is reversed, the field becomes 30 `mu`T. Find the ratio `i_(1)i_(2)`.A. 1B. 3C. 2D. 4 |
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Answer» Correct Answer - c `(mu_(0))/(4pi) (2l_(1))/r-(mu_(0))/(4pi) (2l_(2))/r=10 muT` `(mu_(0)2l_(1))/(4pir) +(mu_(0))/(4pi) (2l_(2))/r=30 muT` On solving, `l_(1)=20 A ` and `l_(2)=10 A` So, `l_(1)//l_(2)=2` |
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| 183. |
A galvanometer having a coil resistance of `60 Omega` shows full scale defection when a current of `1.0 A` passes thoguth it. It can vbe convered into an ammeter to read currents up to `5.0 A` byA. putting in parallel a resistance of `240 Omega`B. putting in series a resistance of `15 Omega`C. putting in series a resistance of `240 Omega`D. putting in parallel a resistance of `15 Omega` |
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Answer» Correct Answer - D To convert a galbonmeter to ammeter to ammrter a small resistance is connected in parallel to the coil of the galvanometer.n Here, `G_(1) = 60 Omega I_(g) = 1.0 A, I = 5A` `I_(R) G_(1) = (1 - 1_(g))S` `S = (I_(g) G_(1))/(I - I_(g)) = (1)/(5 - 1) xx 60 = 15 Omega` Putting `15 Omega` resistance in parallel. |
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| 184. |
A charged particle of mass m and charge q is accelerated through a potential difference of V volts. It enters a region of uniform magnetic field which is directed perpendicular to the direction of motion of the particle. Find the radius of circular path moved by the particle in magnetic field. |
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Answer» Since the particle is accelerated through V volts, therefore it kinetic energy will be equal to qV. or `1/2mv^2=qV`. Therefore, `v=sqrt((2qV)/m)`. Radius of circular path is given by `R=(mv)/(qB)=sqrt((2mV)/(qB^2))`. |
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| 185. |
When a current carrying coil is placed in a uniform magnetic field with its magnetic moment anti-parallel to the field.A. torque on it is maximumB. torque on it is zeroC. potential energy is maximumD. dipole is in unstable equilibrium |
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Answer» Correct Answer - (b,c,d) `vectau=vecMxxvecB and U=-vecM.vecB` Here, `vecM and vecB` are anti-parallel. `:. Vectau=vecO and U=+MB (maximum)` `:.` (b),(c) and (d). |
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| 186. |
Magnetic fieldA. can increase the speed of charged particleB. can accelerate a charged particleC. both a and b are correctD. both a and b are incorrect |
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Answer» Correct Answer - b Magnetic field can accelerate a charged particle by changing the direction of its velocity but it cannot change the speed of charged particle as magnetic force always acts perpendicular to the velocity of charged particle. |
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| 187. |
A solenoid of length 1.0 m has a radius of 1 cm and has a total of 1000 turns wound on it. It carries a current of 5A. If an electron was to move with a speed of `10^(4) ms^(-1)` along the axis of this current carrying solenoid, then force experienced by this electron isA. 2NB. 1.2 NC. zeroD. 2.5 N |
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Answer» Correct Answer - c Here, L=1m , N=1000 The number of turns per unit length n=N/L =1000 turn/m Magnetic field inside the solenoid `B=mu_(0)nI=mu_(0)xx1000xx5=2pixx 10^(-3) T` The direction of magnetic field is along the solenoid. For electron , `q=-e, v =10^(4) ms^(-1)` Magnetic Lorentz force, `F=evB sin 0^(@) =0` as the angle between B and v is `0^(@)` |
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| 188. |
Lorentx force can be calculated by using the formula.A. `vec(F) = q(vec(E) + vec(v) xx vec(B))`B. `vec(F) = q(vec(E) - vec(v) xx vec(B))`C. `vec(F) = q(vec(E) + vec(v) . vec(B))`D. `vec(F) = q(vec(E) xx vec(v) + vec(B))` |
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Answer» Correct Answer - A Lorentz force is given by `vec(F)=vec(F)_(e)+vec(F)_(m)=qvec(E)+q(vec(v)xxvec(B))=q[vec(E)+(vec(v)xxvec(B))]` |
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| 189. |
Statement1: The magnetic filed at the ends of very long current carrying solenoid is half of that at the centre. Statement2: If the solenoid is sufficiently long, the field within it is uniform.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 - B For a solenoid `B_(end)=1/2(B_(I n))`. Also for a long solenoid, magnetic field is uniform within it but this reason is not explaining the statement `(I)` |
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| 190. |
Assertion: A linear solenoid carrying current is equivalent to a bar magnet. Reason: The magnetic field lines of both are same.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 - A The magnetic lines of force due to current carrying straight solenoid is as that of bar magnet. |
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| 191. |
If a particle of charge `10^(-12)` coulomb moving along the`hat(x)`-direction with a velocity `10^(5)m//s` experiences a force of `10^(-10)` newton in `hat(y)`-direction due to magnetic field. Then the minimum magnetic field isA. `6.25xx10^(3)` tesla in `hat(z)`-directionB. `10^(-15)` tesla in `hat(z)`-directionC. `6.25xx10^(-3)` tesla in `hat(z)` directionD. `10^(-3)` tesla in `hat(z)`-direction |
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Answer» Correct Answer - D `F=qvB sin thetaimpliesB=(f)/(qvsin theta)` `B_(min)=(F)/(qv)` (when `theta=90^(@)`) `:. B_(min)=(F)/(qv)=(10^(-10))/(10^(-12)xx10^(5))=10^(-3)tesla` in `hat(z)`-direction. |
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| 192. |
There is no change in the energy of a charged particle moving in a magnetic field although a magnetic force is acting on it . |
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Answer» Correct Answer - 1 The magnetic force acts in a direction perpendicular to the direction of velocity and hence it cannot change the speed of the charged particle. Therefore, the kinetic energy `(=1/2mv^2)` does not change. |
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| 193. |
A wire carrying a current `i` is placed in a uniform magnetic field in the form of the curve `y=asin((pix)/L)0 lexle2L`. The force acting on the wire is A. `(iBL)/(pi)`B. `iBLpi`C. 2iBLD. Zero |
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Answer» Correct Answer - C |
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| 194. |
A charged particle is whirled in a horizontal circle on a frictionless table by attaching it to a string fixed at one pint. If a magnetic field is wsitched on in the vertical direction, the tension in the stringA. will increaseB. will decreaseC. remains sameD. may increase or decrease |
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Answer» Correct Answer - d Depending on the direction of magnetic field, tension may increase or decrease. |
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| 195. |
A charged particle moves with velocity `v` in a uniform magnetic field `vecB`. The magnetic force experienced by the particle isA. Always zeroB. Never zeroC. Zero, if `vecB` and `vecV` are perpendicularD. Zero, if `vecB` and `vecv`are parallel |
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Answer» Correct Answer - D `vecF=q(vecvxxvecB), if vecv||vecB then vecF=0` |
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| 196. |
A particle with charge 2.0 C movess through a uniform magnetic field. At one instant the velocity of the particle is `(2.0hati+4.0hatj+6.0hatk)` m`//`s and the magnetic force on the particle is `(4.0hati-20hatj+12hatk)` N. The x and y components of the magnetic fields are equal. What is `vecB` ? |
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Answer» Correct Answer - `vecB=(-3.0hati-3.0hatj-4.0hatk)T` |
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| 197. |
A charged particle moves in a uniform magnetic field. The velocity of the particle at some instant makes an acute angle with the magnetic field. The path of the particle will beA. A straight lineB. A circleC. A helix with uniform pitchD. A helix with non uniform pitch |
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Answer» Correct Answer - C When particle enters at angle other than `0^(@)` or `90^(@)` or `180^(@)`, path followed is helix. |
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| 198. |
A proton (or charged particle) moving with velocity `v` is acted upon by electric field `E` and magnetic field `B`. The proton will move underflected ifA. E is perpenducular to BB. E is parallel to v and perpendicular to BC. E,B and v are mutually perpendicular and v=E/BD. E and B both are parallel to v |
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Answer» Correct Answer - C In this case `|vecF_(e)|=|vecF_(m)|` and both forces are opposite to each other. |
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| 199. |
A proton and a deuteron both having the same kinetic energy, enter perpendicularly into a uniform magnetic field `B`. For motion of proton and deuteron on circular path or radius `R_(p)` and `R_(d)` respectively, the correct statement isA. `R_(d)=sqrt2R_(p)`B. `R_(d)=R_(p)//sqrt2`C. `R_(d)=R_(p)`D. `R_(d)=2R_(p)` |
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Answer» Correct Answer - A `(mv^(2))/(R)=qvB.` For proton `R_(p)=(mv)/(qB)=sqrt(2m_(p)E)/(qB)` and for deuteron `R_(d)=(sqrt(2m_(d)E))/(qB)` `implies (R_(d))/(R_(p))=sqrt(m_(d)/(m_(p)))=sqrt(2) impliesR_(d)=sqrt(2)R_(p)` |
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| 200. |
An electron is moving along positive x-axis. To get it moving on an anticlockwise circular path in x-y plane, a magnetic field is appliedA. along positive y-axisB. along positive z-axisC. along negative y-axisD. along negative z-axis |
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Answer» Correct Answer - b For the particle to move along anticlockwise path, force should be along `hatj`. Velocity is along `hati`. Now, `vecF_m=-e(vecVxxvecB)` In terms of unit vectors only, `hatj=-(hatixxhat?) or hatj=hat?xxhati` Clearly, `hat? is hatk.` |
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