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Physics NEET MCQ
Magnetic Effects Of Current Mcq
Quiz 2
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Q.1
Charge q is uniformly spread on the conductor radius R. The ring rotates about its axis with uniform frequency f Hz. the magnitude of magnetic field induction at the centre of the ring is .... [ CBSE-PMT 2011]
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a)
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b)
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c)
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d)
Explanation
Current I=q/T=qf Magnetic field at the centre of the ring is Answer: (a)
Q.2
A uniform electric field and uniform magnetic field are acting along the same direction in a certain region. If an electron is projected in the region such that its velocity is pointed along the direction of fields, then the electron ... [ CBSE-PMT 2011]
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a) Will turn towards right of direction of motion
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b) Speed will decrease
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c)Speed will increase
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d)Will turn towards left direction of motion
Explanation
v and B are in same direction so magnetic force on electron become zero, only electric force acts. But force on electron due to electric field is opposite to the direction of velocity Answer:(b)
Q.3
A charged particle having charge q is moving in a circle of radius R with uniform speed v. The associated magnetic moment µ is given by.. [ CBSE-PMT 2007]
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a) qvR2
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b) qvR2 / 2
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c) qvR
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d) qvR/2
Explanation
periodic time T=2πR /v Thus frequency f=v/ 2πR Current I=qf I=qv/ 2πR Magnetic moment µ=IA A=πR2 µ=qvR / 2 Answer: (d)
Q.4
A galvanometer having a coil resistance of 60Ω shows full scale deflection when current of 1.0 amp passes through it. It can be converted into an ammeter to read current up to 5.0amp by.. [ CBSE-PMT 2009]
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a)putting in series a resistance of 15Ω
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b) putting in series a resistance of 240Ω
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c)putting in parallel a resistance of 15Ω
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d)putting in parallel a resistance of 240Ω
Explanation
To convert galvanometer in to ammeter resistance should be connected in parallelShunt resistance formula Here Current capacity of galvanometer, Ig=1 AmpCurrent to be measure I=5 AmpResistance of Galvanometer G=60ΩOn substituting values in above formula we get S=15Ω to be connected in parallelAnswer: (c)
Q.5
A galvanometer has a coil of resistance 100Ω and gives a full-scale deflection for 30 mA current. It is to work as a voltmeter of 30 Volt range, the resistance required to be added will be .... [ CBSE-PMT 2010]
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a) 900 Ω
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b) 1800 Ω
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c)500 Ω
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d)1800 Ω
Explanation
To convert galvanometer resistance should be connected in seriesV=Ig ( G + R) Here G is the resistance of galvanometer=100 ΩCurrent creating of galvanometer Ig=30 mA=30×10-3 Voltage to be measure V=30 V By substituting the values in above equation we get R=900ΩAnswer: (a)
Q.6
An electron enters a region where magnetic field(B) and electric field E are mutually perpendicular , then... [ CBSE-PMT 1994]
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a) It will always move in the direction of B
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b) It will always move in the direction of E
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c)it will always posses circular motion
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d)it can go un deflected also
Explanation
When the deflection produced by electric field is equal to the deflection produced by magnetic field, then the electron can go un deflected.. Answer:(d)
Q.7
A current carrying coil is subject to a uniform magnetic field. The coil will orient so that its plane becomes... [ CBSE-PMT 1988]
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a) inclined at 45° to the magnetic field
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b) inclined at any arbitrary angle to the magnetic field
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c) parallel to the magnetic field
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d) Perpendicular to the magnetic field
Explanation
The plane of coil will itself so that area vector aligns itself along the magnetic field. So, the plane will orient perpendicular to the magnetic field. Answer: (d)
Q.8
A coil of one turn is made of wire of certain length and then from the same length a coil of two turns is made. if the same current is passed in both the cases, then the ratio of the magnetic inductions at their centres will be ... [ CBSE-PMT 1998]
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a)2:1
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b) 1:4
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c)4:1
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d)1:2
Explanation
Let 'l' be the of the wireFirst casel=2πr r=l / 2π Now here n is number of turns per unit length=1 n=2πr / l substituting value of r in above equation we getSecond casenumber of turns are two l=2(2πr') r'=l / 4πOn substituting above values in the formula of magnetic field we getAnswer: (b)
Q.9
Tesla is the unit of... [ CBSE-PMT 1988]
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a) magnetic flux
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b) magnetic field
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c)magnetic induction
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d)magnetic moment
Explanation
Tesla is the unit of magnetic fieldAnswer: (b)
Q.10
The magnetic field at a distance 'r' from a long wire carrying current 'i' is 0.4 tesla. The magnetic field at a distance '2r' is ..... [ CBSE-PMT 1992]
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a) 0.2 tesla
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b) 0.8 tesla
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c)0.1 tesla
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d)1.6 tesla
Explanation
We know that for long wire B∝ (1/r)Thus, when 'r' s doubled, the magnetic field become half, Magnetic field be half of earlier i.e 0.2T Answer:(a)
Q.11
Under the influence of a uniform magnetic field a charged particle is moving in a circle of radius R with constant speed v . The time period of motion... [ CBSE-PMT 2007]
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a) depends on both R and v
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b) is independent of both R and v
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c) depends on R and not on v
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d) depends on v and not on R
Explanation
Since particle is performing under influence of magnetic field . The time period of the motion is independent of R and vAnswer: (b)
Q.12
The magnetic force acting on a charged particle of charge -2µC in a magnetic field of 2T acting in y direction, when the particle velocity is (2i+3j)×106 ms-1, is... [ CBSE-PMT 2009]
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a)4N in z-direction
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b) 8N in y-direction
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c)8N in z-direction
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d)8N in -z-direction
Explanation
The magnetic force acting on the charged particle is given byF=q(v × B )F=-2×10-6 [ (2i+3j)×106 × (2j)]F=-8k ∴ Force is 8N along z-axisAnswer: (d)
Q.13
Two circular coils 1 and 2 are made from the same wire but the radius of 1st coil is twice that of the 2nd coil. What potential difference in volts should be applied across them so that the magnetic magnetic field at the centres is the same... [ CBSE-PMT 2006]
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a) 4 times of first coil
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b) 6 times of first coil
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c)2times of first coil
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d)3times of first coil
Explanation
Since material same, but radius of the wires are different, now resistance depends on lengthlet r1 be the radius of first coillet r2 be the radius of second coilgiven r1=2r2 If R1 is resistance of 1st coil I1 is the current through 1st coilI2 is the current through 2nd coil and R2 is resistance of 2nd coil then R1=2R2Let V1 be the potential across 1stletV2 be the potential across 2nd coil Now I1=V1 / R1 and I2=V2 / R2 Now magnetic field at the centre of coil is given by formulaNow to have same magnetic field at centre Answer: (a)
Q.14
Total charge induced in a conducting loop when it is moved in magnetic field depends on... [ CBSE-PMT 1990]
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a) the rate of change of magnetic flux
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b) initial magnetic flux only
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c)the total change in magnetic flux
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d)final magnetic flux only
Explanation
We know that induced voltage ε=dΦ / dt if coil is connected to resistance R then ε=i R thus Total charge induced q=∫ i dt If Φ1 and Φ2 is final and initial flux linked with coil Answer:(c)
Q.15
A galvanometer of 50 Ω resistance has 25 divisions. A current of 4×10-4 ampere gives a deflection of one per division. To convert this galvanometer into voltmeter having a range of 25 volts, it should be connected with a resistance of ... [ CBSE-PMT 2004]
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a) 2450 Ω in series
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b) 2500 Ω
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c) 245 Ω
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d) 2550 Ω
Explanation
Current capacity of galvanometer Ig=25 ×4×10-4=10-2 Resistance of galvanometer G=50 Ω Range of voltmeter=25 V From formula resistance to be connected in series R=(V/Ig - G R=2450 Ω Answer: (a)
Q.16
A deuteron of kinetic energy 50keV is describing a circular orbit of radius 0.5 metre in a plane perpendicular to the magnetic field B. The kinetic energy of proton that describes a circular orbit of radius 0.5 metre in the same plane with same B is .. [ CBSE-PMT 1991]
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a)25 keV
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b) 50 keV
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c)200 keV
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d)100 keV
Explanation
We know that Now If mp is mass of proton and md mass of deuteronEp is kinetic energy of proton and Ed is kinetic energy of deuteronThen Ep / Ed=md / mpWe know that mass of deuteron in twice the mass of proton thus Ep / Ep=2 Ep=2×Ed Ep=2×50=100 keVAnswer: (d)
Q.17
Two long parallel wires are at a distance of 1 metre. Both of them carry one ampere of current. The force of attraction per unit length between the two wires is ... [ CBSE-PMT 1998]
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a) 2 ×10-7 N/m
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b) 2 ×10-8 N/m
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c) 5 × 10-8 N/m
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d)10-7 N/m
Explanation
Force between the current carrying wire is given byAnswer: (a)
Q.18
A straight wire of length 0.5 metre and carrying a current of 1.2 amp is placed in uniform magnetic field of induction 2 tesla. The magnetic field is perpendicular to the length of the wire. The force on the wire is ... [ CBSE-PMT 1992]
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a) 2.4 N
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b) 1.2 N
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c)3.0 N
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d)2.0 N
Explanation
F=Bil F=2×1.2×0.5=1.2 N Answer:(b)
Q.19
A charged particle moves through a magnetic field in a direction perpendicular to it. Then the .. [ CBSE-PMT 2003]
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a) velocity remains unchanged
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b) speed of the particle remains unchanged
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c) direction of the particle remains unchanged
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d) acceleration remains unchanged
Explanation
Magnetic force acts perpendicular to the velocity. Hence speed remains constant.Answer: (b)
Q.20
A galvanometer of resistance, G is shunted by a resistance S ohm. To keep the main current in the circuit unchanged, the resistance to be put in series with the galvanometer is ... [ CBSE-PMT 2011]
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a)
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b)
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c)
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d)
Explanation
To keep the main current in the circuit unchanged, the resistance of the galvanometer should be equal to the resistanceAnswer: (c)
Q.21
In mass spectrometer used for measuring the masses of ions, the ions are initially accelerated by an electric potential V and then made to describe semicircular path of radius R using magnetic field B. If V and B are kept constant, the ratio charge/ mass will be proportional to ... [ CBSE-PMT 2007]
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a) 1/ R2
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b) R2
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c)R
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d)1/R
Explanation
In mass spectrometer, when ions are accelerated through potential V ½ ( m v2 )=qV --eq(1) As the magnetic field curves the path of the ions in a semicircular orbit Bqv=mv2 / R v=BqR / m ---eq(2) Substituting value of eq(2) in eq(1) since V and B are constants,∴ q/m ∝ 1/R2 Answer: (a)
Q.22
A square loop, carrying a steady current I, is placed in a horizontal planer near a long straight conductor carrying a steady current I1 at a distance 'd' from the conductor as shown in figure. The loop will experience .. [ CBSE-PMT 2011]
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a) a net repulsive force away from the conductor
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b) a net torque acting upward perpendicular to the horizontal plane
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c)a net torque acting downward normal to the horizontal plane
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d) a net attractive force towards the conductor
Explanation
Force F ∝ 1/dForce on left and right side of loop is equal and opposite Force on the side parallel to wire is opposite to each other But side near to wire will be more than the other parallel side Hence the net attraction force will be towards the conductor. Answer:(d)
Q.23
Four wires each of length 2.0metres are bent into four loops P, Q, R and S and then suspended into uniform magnetic field. Same current is passed in each loop. Which statement is correct [ MPPMT 1995]
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a) Couple on loop P will be the highest
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b) Couple on loop Q will be the highest
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c)Couple on loop R will be the highest
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d)Couple on loop S will be the highest
Explanation
τ ∝ area of loopAnswer: (d)
Q.24
A current loop consists of two identical semicircular parts each of radius R, one lying in x-y plane and the other in x-z plane. If the current in the loop is 'i', the resultant magnetic field due to the two semicircular parts at their common centre is ... [ CBSE-PMT 2010]
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a)
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b)
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c)
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d)
Explanation
Magnetic field at the centre are respectivelySince both fields are perpendicular to each other resultant field will be Answer: (b)
Q.25
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 force acting on the loop is [ IIT 1983]
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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
the magnetic field is perpendicular to the plane of the paper. Let us consider two diametrically opposite element. By Fleming left hand rule. We find the force is in opposite direction and are in same plane and line of action of the forces are passing through same point centre of circle , similarly, force acting on diametrically opposite elements will cancel out in pair. Thus net force is zeroAnswer: (c)
Q.26
A battery is connected between two points A and B on the circumference of a uniform conducting ring of radius r and resistance R. One of the arc AB of the ring subtends as angle θ at the centre. The value of the magnetic induction at the centre due to the current in the ring is [ IIT 1995]
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a) proportional to 2(180° - θ)
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b) inversely proportional to r
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c)zero, only if θ=180°
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d)zero for all values of θ
Explanation
If ρ is resistance of wire per unit length then resistance of arc ABC=Rθρ Current through arc ACB I1=E / (Rθρ) Magnetic field at centre due to arc ABC, directed upward=B1= resistance of arc ACB=R(2π-θ)ρ Current through arc ADB I1=E / [R(2π-θ)ρ] Magnetic field at centre due to arc ADB, directed downward=B2= From above it is clear that B1 and B2 are equal and opposite in direction thus resultant magnetic field is zeroAnswer: (d)
Q.27
A proton, a deuteron and an α-particle having the same kinetic energy are moving in circular trajectories in a constant magnetic field. If rp, rd and rα denote respectively the radii of the trajectories of these particles, then [ IIT 1999]
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a) rα=rp < rd
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b) rα > rd >rp
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c)rα=rd >rp
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d)rα=rd=rp
Explanation
particles are moving in circular path thus Centripetal force=magnetic force Kinetic energy E=p2 / 2m p=√(2Em)qBr=√(2Em)Since E ,B are constant r ∝ √m / q ∴rp : rd : rα=√1 / 1 : √2/1 : √4/1 ∴rp : rd : rα=1 : √2 : 1 Thus rα=rp < rd Answer:(a)
Q.28
A circular loop of radius R, carrying current I, lies in X-Y plane with its centre at origin. The total magnetic flux through X-Y plane is [ IIT 1999]
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a) directly proportional to I
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b) directly proportional to R
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c) Inversely proportional to R
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d) zero
Explanation
The magnetic lines of force created due to current will be in such a way that on X-Y plane these lines will be perpendicular. Further, these lines will be in circular loops. The number of line smoving downwards in X-Y plane will be same in number to that coming upwards of X-Y plane. Therefore, the net flux will be zero. One such magnetic line is shown in the figure Answer: (d)
Q.29
A charged particle is released from rest in a region of steady and uniform electric and magnetic field which are parallel to each other. the particle will move is a [ IIT 1999]
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a)straight line
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b) circle
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c)helix
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d)cycloid
Explanation
Direction of motion of charged particle is in direction of magnetic field no force will be exerted by magnetic filed Direction of charge particle is in direction of electric field thus either force will be in the direction of motion of charged particle or opposite depending on the charge but charged particle will move in straight lineAnswer: (a)
Q.30
A particle of charge q and mass m moves in a circular orbit of radius r with angular speed ω. The ratio of the magnitude of its magnetic moment to that of its angular momentum depends on [ IIT 2000]
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a) ω and q
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b) ω m and q
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c)m and q
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d) ω and m
Explanation
The angular momentum L of the particle is given by L=mr2ω where ω=2πf ∴ frequency f=ω / 2πCurrent i=qf=ωq / 2π Magnetic moment M=iA=(ωq / 2π) × πr2 ∴ M=ωqr2 / 2So M/L=ωqr2 / 2mωr2=q/2mAnswer: (c)
0 h : 0 m : 1 s
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