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Physics NEET MCQ
Magnetic Effects Of Current Mcq
Quiz 6
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Q.1
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.2
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.3
In the hydrogen atom, the electron revolves in circular orbit of radius 0.53×10-10 metre and makes 6.6×1015 r.p.s. Then the magnetic dipole moment is approximately [ MPPMT 1999]
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a)10-29 amp×metre2
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b) 10-27 amp×metre2
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c)10-23 amp×metre2
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d)10-19 amp×metre2
Explanation
Current due to revolution=Q/T=e/Tperiodic time T=1/ number of revolution per second1 /T=number of revolution per second Area of circular loop=π r2A=0.88×10-20Now magnetic dipole moment µ=iA Answer: (c)
Q.4
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.5
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.6
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.7
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.8
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.9
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.10
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.11
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.12
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.13
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.14
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)
Q.15
An electron of mass 9×10-31 kg, charge 1.6 ×10-19 C moving with velocity 106 m/s enters a region of magnetic field of strength 5×10-5T perpendicular to the field. Find the radius of the circle described by it
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a) 9.0 cm
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b) 11.25 cm
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c) 13.5 cm
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d) 15.75 cm
Explanation
use formula r=mv/qB Answer: (b)
Q.16
A proton of mass 1.67×10-27 kg and charge 1.6×10-19 C is projected with a speed of 2×106 m/s at an angle of 60° to the x-axis. If a uniform magnetic field of 0.104 tesla is applied along the Y-axis, the path of the proton is [ IIT 1995]
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a)a circle of radius=0.2 m and time period=π×10-7 s
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b) a circle of radius=0.1 m and time period=2π×10-7 s
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c)a helix of radius=0.1 m and time period=2π×10-7 s
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d)a helix of radius=0.2 m and time period=4π×10-7 s
Explanation
Since charged particle is projected, making angle of 60° with the field direction, the path of the particle is a helix. Component of velocity perpendicular to magnetic field causes circular motion.The radius of the helical path is Time period T Answer: (c)
Q.17
A current is flowing in a circular loop of radius R and the magnetic field at its centre is Bo. At what distance from the centre on the axis of the coil, the magnetic field will be Bo / 8? [ MPPMT 1997]
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a) R√7
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b) R√3
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c)2R
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d)8R
Explanation
Answer: (b)
Q.18
Three long, straight and parallel wires carrying currents are arranged as shown in figure. the wire C which carries a current of 5.0 amp is so placed that it experiences no force. the distance of wire C from D is then [ AMU 1995]
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a) 9 cm
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b) 7 cm
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c)5 cm
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d) 3 cm
Explanation
Force between current carrying parallel wires is given by formulaForce on wire C due to A=Force on C due to wire B Answer:(a)
Q.19
A conductor in the form of right angle ABC, with AB=3 cm and BC=4 cm, carries a current of 10A. There is uniform magnetic field of 5T, perpendicular to the plane of the conductor. The force on the conductor AC will be [ MPPMT 1997]
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a) 1.5 N
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b) 2.0 N
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c) 2.5 N
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d) 3.5 N
Explanation
Given AB=3 cm and BC=4 cm ΔABC is right angled triangle thus AC=5cm Force on AC=Bil=5×5×10-2=2.5 N Answer: (c)
Q.20
Along vertical wire in which current is flowing produces a neutral point with the earth's magnetic field at distance of 5 cm's from the wire. If the horizontal component of the earth's magnetic induction is 0.18 gauss then current in the wire is
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a) 0.45 ampere
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b) 4.5 ampere
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c)0.9 ampere
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d)0.036 ampere
Explanation
Answer: (b)
Q.21
A very long solenoid has 800 turns per metre length of solenoid. A current of 1.6 ampere flows through it. Then the magnetic induction at the middle point of the solenoid on its axis, is approximately
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a) 16 × 10-4 tesla
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b) 8 × 10-4 tesla
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c)32 × 10-4 tesla
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d)4 × 10-4 tesla
Explanation
Use formula µonI Answer:(a)
Q.22
P is a point at a distance a from a long, thin straight wire carrying current i as shown in figure. The magnetic field at P is
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a) µoi / 2πa
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b) µoi / 4πa
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c) µoi / 2a
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d) µoi / 4a
Explanation
Put θ1=0° and θ2=90° Answer: (b)
Q.23
A long thick metallic cylinder of radius R has a current of i ampere uniformly distributed over its circular cross-section. then, magnetic induction B away from the axis at a distance r from the axis varies as shown in figures given in options
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a)
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b)
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c)
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d)
Explanation
Answer: (d)
Q.24
A cable carries a current of 1A vertically upward. The magnetic field produced by it at a point 10 cm north will be
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a) 2 × 10-6 Tesla west
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b) 2 × 10-6 Tesla east
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c)2 × 10-8 Tesla west
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d)2 × 10-8 Tesla vertically upward
Explanation
Use formula B=(µo/4π) ( 2I/r) Use right hand thumb rule to determine directionAnswer: (a)
Q.25
Magnetic field at the centre of the cube of edge of length a is
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a)zero
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b) 8µo i / a √2
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c)(8µo i√2) / a
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d)6µo i / a √2
Explanation
zero because current in opposite sides are equal and produces equal and opposite magnetic field at the centre Answer:(a)
Q.26
A straight conductor of length 0.3 metre is placed in uniform magnetic field of induction 8 × 10-4 tesla, normal to the lines of the force. A current of 2.5 ampere flows through it. The mechanical force acting on the conductor, is
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a) 6 Newton
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b) 6 × 10-4 newton
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c) 6 × 10-2 newton
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d) 6 × 10-3 newton
Explanation
F=I l × B given angle is 90° F=IlB on substituting we get option "b" Answer: (b)
Q.27
A circular wire is carrying a current in anticlockwise direction as shown in figure . A long infinite wire CD is placed above the plane of page and carries a current as shown. The force acting on wire CD is
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a)perpendicular to plane if page and directed towards reader
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b) perpendicular to plane of page and directed away from reader
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c)perpendicular to wire CD, in the plane of page and directed towards left of wire
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d)perpendicular to wire CD in the plane of page and directed to the right of wire CD
Explanation
Answer: (c)
Q.28
A long wire carries a steady current. First it is bent into a circular coil of one turn when magnetic induction at the centre is B. Then the same wire is bent to form a circular coil of smaller radius but n turn when the magnetic induction at the centre is B', then
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a) B'=B
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b) B'=n B
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c)B'=n2B
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d)B=n2B'
Explanation
let L be the length of wire L=2πR ∴ R=L/2π If the wire is bent in n loop then L=n ×2πr or r=L/n2π Answer: (c)
Q.29
A rectangular coil of area A of N turns has a current I flowing in clockwise direction when looked at from above. The magnetic moment associated with it
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a) points upwards
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b) points vertically downwards
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c)is zero
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d)is directly proportional to A2
Explanation
Answer:(b)
Q.30
Two concentric circular coils 1 and 2 have radii 20cm and 10cm respectively lie in the same plane. the current in coil 1 is 0.5A in anticlockwise direction. The current in coil 2, so that net field at the common centre is zero, is
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a) 0.5 A in anticlockwise direction
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b) 0.25 A in anticlockwise direction
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c) 0.25 A in clockwise direction
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d) 0.125 A in clockwise direction
Explanation
both coils should produce magnetic field of equal and opposite in directionfrom the formula for magnetic field at centre of coil Answer: (c)
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