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Physics NEET MCQ
Quiz 5
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Q.1
A electron is moving along positive x-axis. To get it move on an anticlockwise circular path in x-y plane, a magnetic filed is applied [ MPPMT 1999]
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a) along positive y-axis
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b) along positive z-axis
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c)along negative y-axis
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d)along negative z-axis
Explanation
As the electron have moved in anticlockwise direction force on electron is on upward direction it is long +y axis. By Fleming's left hand rule we get direction magnetic field is along positive z-axis Answer:(b)
Q.2
Two wires A and B carry currents as shown in figure the magnetic interactions:
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a)push I2 away from I1
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b) pull I2 closer to I1
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c)Turn I2 clockwise
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d)Turn I2 counter clockwise
Explanation
Magnetic field produced by wire A in perpendicular to paper outwards. and lower part of wire it is perpendicular going inward. From Fleming's left hand rule force on wire B is towards positive x-axis for upper part while along negative axis for lower part, thus wire B will turn clock wiseAnswer: (c)
Q.3
The meniscus of liquid contained in one of the limbs of a narrow U tube is placed between the pole pieces of an electromagnet with the meniscus in a line with the field. the liquid is seen to rise to line. This indicates that the liquid is [ MNR 1992]
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a) ferromagnetic
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b) paramagnetic
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c)diamagnetic
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d)non-magnetic
Explanation
Paramagnetic substance move from weak magnetic field to strong magnetic fieldAnswer: (b)
Q.4
A circular loop of area 0.01 m2 and carrying a current of 10amp is placed perpendicular to a magnetic field of intensity of 0.1 tesla. The torque ( in Nm) acting on the loop is [ CBSE 1994]
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a)1.1
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b) 0.8
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c)0.001
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d)0.01
Explanation
τ=NIABsinθτ=(1)(10)(0.01)(0.1)sin90=0.01 Nm Answer:(d)
Q.5
A current of 10 amp is flowing in a wire of length 1.5 metre. A force of 15 N acts on it when it placed in a uniform magnetic field of 2 tesla. the angle between the magnetic field and the direction of the current is [ MPPMT 1994]
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a) 30°
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b) 45°
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c) 60°
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d) 90°
Explanation
F=Il×B F=ILB sinθ 15=(10)(1.5)(2) sinθ sinθ=1/2 θ=30° Answer: (a)
Q.6
There will be a force of repulsion between [ ISM Dhanbad 1994]
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a)two parallel streams of electrons moving in the opposite direction
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b) two parallel wires carrying current in the opposite direction
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c)two parallel electron streams going in the same direction
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d)two parallel wires carrying current in the same direction
Explanation
Answer: (a, b)
Q.7
The magnetic field at a point at a large distance x on the axis of a current carrying circular coil of small radius is proportional to [ EAMCET 1987]
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a) x2
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c)x3
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b) 1/x²
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d)1 / x³
Explanation
From the formula for magnetic field at very large distance form the coil having radius R is Answer: (d)
Q.8
A wire of length L metre carrying current i amp is bent in the form of circle. The magnitude of magnetic moment is [ MPPMT 1995]
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a) iL2 / 4π
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b) iL2 / 2π
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c)4π2iL2
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d)πL2 i
Explanation
circumference of loop=length 2πr=L r=L / 2π Area of loop=πr2=L2 /4π Magnetic moment µ=NIA µ=iL2 /4π Answer:(a)
Q.9
Two straight long conductors AOB and COD are perpendicular to each other and carry currents I1 and IThe magnitude of the magnetic induction at a point P at a distance a from O in a direction perpendicular to the plane ABCD is [ MPPMT 1994]
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a)
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b)
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c)
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d)
Explanation
The point P lies above the plane of paper at a distance a from it. Field at P due to conductor AOB is towards right ( east) Field at P due to COD is tword south is There for net field at P Answer: (c)
Q.10
A current carrying rectangular coil is placed in a uniform magnetic field. In which orientation, the coil will not rotate? [ PMT 199]
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a)the magnetic field is perpendicular to the plane of the coil
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b) the magnetic field is parallel to the plane of coil
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c)the magnetic field is at 45° with the plane of the coil
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d)always in any orientation
Explanation
Torque τ=NABsinθ θ is the angle between area vector and magnetic field When coil is perpendicular to magnetic filed area vector becomes parallel to magnetic field thus θ=0 . and sin0=0Answer: (a)
Q.11
A charge +Q is moving upwards vertically. It enters a magnetic field directed to the north. The force on the charge will be towards [ PMT 1995]
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a) North
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b) South
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c)East
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d)West
Explanation
Answer: (d)
Q.12
A current carrying circular loop is freely suspended by a long thread. The plane of the loop will point in the direction [ PMT 1995]
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a) where ever left free
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b) north south
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c)east-west
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d)at 45° with the east-west direction
Explanation
A current carrying circular loop acts as a bar magnet Answer:(c)
Q.13
An electric charge in uniform motion produces......
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a) an electric field only
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b) a magnetic field only
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c)both electric and magnetic field
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d)no field at all
Explanation
Answer: (c)
Q.14
A conducting loop of radius 'a' carries a constant current I. It is placed in a uniform magnetic field B such that B is perpendicular to the plane of loop. ThE magnetic force acting on the loop is
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a) IaB
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b) 4πaIB
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c)zero
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d)πaIB
Explanation
Answer:(c)
Q.15
A straight thin conductor is bent as shown in figure. It carries a current i amp. the radius of the circular arc is r metre, then the magnetic induction at centre of the semicircular arc is [ MPPMT 1998]
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a) zero
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b) α
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c)
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d)
Explanation
Magnetic induction due to straight part of the conductor is zero Magnetic field at the centre of the loop is given by µonI /2r Here n is the number of turns For 2π=1 turn Thus for π, number of turns=π / 2π Answer: (d)
Q.16
A circular coil of radius R is placed in a uniform magnetic field such that the plane of coil is perpendicular to the magnetic field. If a current i passes through the coil, the torque acting on the coil is [ CPMT 1993]
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a)BiπR2
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b) Bi( πR2/ 2)
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c)BiπR2 / √2
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d)zero
Explanation
B is parallel to A ∴ τ=0Answer: (d)
Q.17
When a stationary charged particle is placed near a stream of moving charges, then the stationary charged particle [ CPMT 1993]
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a) will experience no force
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b) will experience a force due to electric field only
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c)will experience a force due to magnetic field only
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d)will experience a force due to electric and magnetic fields both
Explanation
Moving charges produce electric and magnetic field both. But magnetic field can not exert any force on stationary chargeAnswer: (b)
Q.18
A current i amp flows along an infinitely long straight conductor. If r metre is the perpendicular distance of a point from the lower end of the conductor, then the magnetic induction b is given by [ MPPMT 1994]
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a)
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b)
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c)
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d)
Explanation
Magnetic field due to finite conductor isat one end of the conductor of finite length θ1=0 and other end θ2=90° Answer:(b)
Q.19
The velocity of helium nucleus traveling in the a current path in a magnetic field is v. The velocity of the proton moving along the same path in the same magnetic field field is [ CPMT 1993]
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a) 4v
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b) 2v
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c) v
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d) v/2
Explanation
From the formula for momentum p=qBr radius is same in both the cases Thus for proton mp vp=qBr --eq(1) For Helium nucleus mH vH=2qBr But mass of Helium=4 mass of proton and charge on Helium is 2× charge on proton thus 4mp vH=2qBr --(eq2) Taking ratio of equation 1 and equation 2 we get Answer: (b)
Q.20
Which of the following graph shows the variation of magnetic induction B with distance r from a long wire carrying current : [ MPPMT 1999]
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a)
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b)
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c)
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d)
Explanation
Magnetic field due to very long conductor ∝ (1/r) which is described by graph c .Answer: (c)
Q.21
A deuteron of kinetic energy 50 KeV is describing a circular orbit of radius 0.5 metre in a plane perpendicular to magnetic field B. The kinetic energy of the proton that describe a circular orbit of radius 0.5 metre in the same plane with the same B is [ CBSE 1991]
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a) 25KeV
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b) 50KeV
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c)200KeV
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d)100KeV
Explanation
Form the formula for momentum p=qBr But p=√(2mE) ∴ 2mE=(qBr)2 radius and charge in both the cases is same but mass of deuteron is=2× mass of proton thus 2mpEp=2mdEd2mpEp=4mp50Ep=2×50=100Kev Answer:(d)
Q.22
A uniform magnetic field acts at right angles to the direction of motion of electrons. As a result, the electron moves in a circular path of radius 2 cm. If the speed of the electrons is doubled, then the radius of the circular path will be [ CBSE 1991]
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a) 2.0 cm
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b) 0.5 cm
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c) 4.0 cm
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d) 1.0cm
Explanation
From the formula of momentum P=qBr mv=qBr Thus r ∝ v . If velocity is doubled radius is doubled new radius=4 cm Answer: (c)
Q.23
A current loop is placed in a uniform magnetic field. The loop will experience [ CBSE 1993]
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a)zero linear force but may experience a torque
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b) a linear force and may experience a torque also
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c)A linear force only
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d)zero linear force but will necessarily experience torque
Explanation
Resultant force on loop will be zero. If area vector of loop is parallel to magnetic field torque will be zero else it will be non zeroAnswer: (a)
Q.24
A particle of charge q and mass m moving with velocity v along the x-axis enters the region x > 0 with uniform magnetic field B along the k direction. The particle will penetrate in this region in the x-direction up to a distance d equal to [ MPPMT 1997]
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a) zero
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b) mv/qB
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c)2mv/qB
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d)infinity
Explanation
From the formula for momentum mv=qBr particle follows circular path hence maxium distance will be diameter of path radius=mv/qB diameter=2mv/ qBAnswer: (c)
Q.25
A current of 1 amp is passed through a straight wire of length 2.0 metres. The magnetic field at a point in air at distance of 3 metres from either end of wire and lying on the axis of wire will be [ MPPMT 1995]
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a) µo / 2π
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b) µo / 4π
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c)µo / 8π
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d)zero
Explanation
Point on the axis of wire hence zero magnetic field Answer:(d)
Q.26
A wire of fixed length L can be formed into many circular loops of varying radii r depending on the number of turns n. The loop so formed is carrying a current and is so placed normally in a uniform magnetic field B. In order that the torque on the circular loop formed be maximum, the number of turns n must be equal
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a) 1
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b) 4
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c) 8
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d) ∞
Explanation
LEt N be the turns in the loop L=N(2πr) r=L / (2Nπ) Torque τ=NIAB for τ maximum, n should be minimum. Minimum value of N=1 Answer: (a)
Q.27
A proton and an alpha particle enter in a uniform magnetic field with same velocity. The period of rotation of the alpha particle will be [ mPPMT 1990]
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a)four times that of proton
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b) two times that of proton
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c)three times that of proton
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d)same as that of the proton
Explanation
Momentum p=qBr mv=qBr but v=ωr thus mωr=qBr mω=qB 2π /T=qB / m T=m2π / qB Thus T ∝ m/q mass of alpha particle is four times the mass of protonand charge on alpha particle is two time the charge on proton thus time period period of alpha particle is two times of the protonAnswer: (b)
Q.28
An electron of charge e is going around in an orbit of radius R metres in hydrogen atom with velocity v m/s. The magnetic flux density associated with it at its centre is [ CBSE 1993]
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a)
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b)
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c)
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d)
Explanation
Formula for magnetic field at centre due charge revolving in circular orbit is Answer: (a)
Q.29
H+, He++ and O++ all having same kinetic energy pass through a region in which there is a uniform magnetic filed perpendicular to their velocity. The masses of H+,He++ and O++ are 1 amu. 4 amu and 16 amu respectively. Then [ IIT 1994]
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a) H+ will be deflected most
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b) O++ will be deflected most
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c)He + and O++ will deflect equally
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d)All will be deflected equally
Explanation
We know that momentum p=qBr and p=√(2mE) Thus √(2mE)=(qBr)r=√(2mE) / qB For H+ rh=√(2E) / eB For He+ rhe=√(8E) / eB For O++ rhe=√(16E) / 2eB rhe=√(8E) / eB From above it is clear that radius of He+ and O++ are same Answer:(c)
Q.30
Weber ampere per meter is equal to [ MPPMT 1990]
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a) joule
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b) newton
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c) henry
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d) watt
Explanation
force on current carrying conductor n magnetic field F=Bil Unit of B in equation is tesla=Weber / Area Answer: (b)
0 h : 0 m : 1 s
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