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Physics NEET MCQ
Magnetic Effects Of Current Mcq
Quiz 5
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Q.1
A coil of 100 turns and area 5 square centimetre is placed in a magnetic field B=0.2T. The normal to the plane of the coil makes angle of 60° with the direction of the magnetic field. The magnetic flux linked with the coil is [ MPPMT 1997]
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a) 5×10-3 Wb
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b) 5×10-5 Wb
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c) 10-2 Wb
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d) 10-4 Wb
Explanation
Φ=NBAcosθ Φ=100×0.2×(5×10-4) ×cos60Φ=5×10-3 Wb Answer: (a)
Q.2
A current carrying loop is placed in a uniform magnetic field. the torque acting on it, does not depend upon [ Raj. PMT 1997]
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a)shape of loop
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b) area of loop
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c)value of current
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d)magnetic field
Explanation
Answer: (a)
Q.3
A straight wire of length 0.5 metre and carrying a current of 1.2 amp. is placed in a uniform magnetic field of induction 2 tesla. The magnetic field is perpendicular to the length of the wire. The force on the wire is: [ BHU 1998]
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a) 2.4N
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b) 1.2 N
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c)3.0N
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d)2.0N
Explanation
F=BilF=2×1.2×0.5F=1.2 NAnswer: (b)
Q.4
A vertical straight conductor carries a current vertically upwards. A point P lies to the east of it at a small distance and another point Q lies to the west at the same distance. The magnetic field at P neglecting earth's field is [ MNR 1988]
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a) greater than at Q
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b) same as at Q
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c)lesser than at Q
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d)greater or lesser than that at Q, depending upon the strength of current
Explanation
there is no magnetic component of earth along east and west thus magnetic field field produced at P and Q is same Answer:(b)
Q.5
Two thin wire carrying equal current are held perpendicular to each other, as shown. Now, AB and CD are perpendicular to each other and symmetrically placed with respect to the current. The resultant magnetic field would be zero [ MPPMT 1995]
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a) on AB
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b) on CD
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ac) on both AB and CD
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d) on OB and OC
Explanation
Magnetic field on line AB is opposite direction, according to right hand thumb rule. Answer: (a)
Q.6
A proton is moving with velocity v in a direction opposite to the magnetic field B. The magnetic force experienced by the proton is .....
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a) Bev
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b) - Bev
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c)Bv
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d)zero
Explanation
Answer: (d)
Q.7
A straight conductor carrying a direct current i amp is split into circular loop as shown in figure. Then the magnetic induction at the centre of the circular loop of radius r metre is [ MPPMT 1997]
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a)0
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b) ∞
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c)µoi / 2πr
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d)µoi / 2r
Explanation
Straight wire is at the centre of loop hence no magnetic induction at centre due to long wire Current direction in the half of loop are parallel to each other hence will cancel out each other effect. Thus net magnetic induction at centre is zeroAnswer: (a)
Q.8
A conducting circular loop of radius r carries a constant current i. It is placed in a uniform magnetic field Bo such that Bo is perpendicular to the plane of the loop. the magnetic fore acting on the loop is [ MPPMT 1999]
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a) irBo
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b) 2πirBo
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c) zero
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d)πirBo
Explanation
From figure, by applying Fleming's left hand rule, we find that current element dl which are opposite to each other, experiences forces in opposite directions are passing through the same point thus resultant force is zero Answer: (c)
Q.9
A direct current is sent through a helical spring The spring [ MPPMT 1998]
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a) tends to get shorter
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b) tends to get longer
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c)tends to rotate about the axis
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d)tends to move northward
Explanation
Turns are parallel and current is in same direction , there will be attractive force between helical turns. so spring will contract and tended to get shorter Answer:(a)
Q.10
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 2cm. 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) 1.0 cm
Explanation
We kno tat centripetal force=magnetic force mv2 / r=qvb Thus v ∝ r When speed of electron becomes double radius becomes double=4.0 cm Answer: (c)
Q.11
A helium nucleus makes a full rotation in a circle of radius 0.8 metre in two seconds. The value of the magnetic field B at the centre of circle will be [ CPMT 1988]
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a)10-19 / µo
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b) 10-19 µo
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c)2×10-19 µo
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d)2×1019 µo
Explanation
When Helium nucleus makes a full rotation. We may consider it as current in loop so magnetic field at the centre of loop B=µoni / 2r but i=q /t i=2e/2=e Answer: (b)
Q.12
A straight section PQ of a circuit lies along the x-axis from x=-(a/2) to x=+(a/2) and carries a steady current i. The magnetic field due to the section PQ at a point x=+a will be [ MPPMT 1987]
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a) proportional to a
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b) proportional to a2
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c)proportional to ( 1 /a)
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d)equal to zero
Explanation
point x=+a is in line of current thus magnetic field will be zeroAnswer: (d)
Q.13
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.14
A proton enters a magnetic field of flux density 1.5 weber/metre2 with a velocity of 2×104 metre / sec at an angle of 30° with the field. The force on the proton will be [ MPPMT 1994]
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a) 2.4×10-12 N
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b) 0.24×10-12 N
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c) 24×10-12 N
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d) 0.024×10-12 N
Explanation
F=qvB sinθ F=1.6×10-16 ×2×104 ×1.5×sin30 F=2.4×10-12 Answer: (a)
Q.15
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.16
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.17
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.18
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.19
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.20
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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b) 1/x2
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c)x3
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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.21
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.22
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.23
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.24
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.25
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.26
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.27
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.28
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.29
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.30
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)
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