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Physics NEET MCQ
Quiz 2
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Q.1
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.2
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.3
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.4
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.5
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.6
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.7
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.8
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.9
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.10
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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d)10-7 N/m
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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
Explanation
Force between the current carrying wire is given byAnswer: (a)
Q.11
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.12
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.13
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.14
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.15
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.16
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.17
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.18
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.19
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.20
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.21
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.22
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.23
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)
Q.24
Two long parallel wires are at a distance 2d apart. They carry steady equal currents flowing out of the plane of the paper,as shown. The variation of the magnetic field B along the line XX' is given by [ IIT 2000]
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a)
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b)
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c)
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d)
Explanation
The wire A and B are perpendicular to the plane of paper and current is towards the reader. Let us consider certain points. Point C : Point C is exactly between the two wires .The magnetic filed at C sue to A ( BCA) is in upward direction but magnetic field at C due to B is in down ward direction. NEt field is zero.Point E : Point E is closed to wire B. Magnetic filed due to A is upward direction and Magnetic filed due to B is in downward direction but |BEA| < |BEB| Thus magnetic field will be in upward directionPoint D: Which is closed to wire A. Magnetic filed due to A is upward direction and Magnetic filed due to B is in downward direction but |BDA| > |BDB| Thus magnetic field will be in upward directingPoint M : Point M is left of wire A : Here magnetic filed due to wire A and B is in same direction and is in downward.Point N : Point N is right side of wire B . HEre magnetic filed due to both the wires is in upward direction From given option option b is correct representation Answer:(b)
Q.25
An infinitely long conducting wire PQR is bent to form a right angle as shown in figure. A current I flows through PQR. The magnetic field due to the current at the point M is HNow another infinitely long straight conductor QS is connected at Q so that current is I/2 in QR as well as QS, the current in PQ remains unchanged. The magnetic field at M is now HThe ratio H1 / H2 is given by [ IIT 2000]
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a) 1/2
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b) 1
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c) 2/3
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d) 2
Explanation
Case I : At point M magnetic filed will not be produced due to QR as pont is on the line So magnetic field at M will b only due to PQ given by Here R is perpendicular distance of point M from PQ Case II: When wire QS is joined direction of current through QS is same as direction of current through PQ Magnetic filed at M H2=Magnetic field due to PQ + Magnetic filed due to QS Thus H1 / H2=2/3Answer: (c)
Q.26
An ionized gas contains both positive and negative ions. If it is subjected simultaneously to n electric file along the +x-direction and magnetic filed along +Z-direction then [ IIT 2000]
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a)positive ions deleted towards +y-direction and negative ions towards -y direction
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b) all ions deflect to wards +y-direction
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c)all ions deflect to wards -y-direction
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d)positive ions deleted towards -y-direction and negative ions towards +y direction
Explanation
Positively charged particle :a)Force due to electric field : in positive X directionb) Force due to magnetic field : This force will move the positively charged particle to wards Negative Y axis Negatively charged particle: a)Force due to electric field : in negative X directionb) Force due to magnetic field : This force will move the positively charged particle to wards Negative Y axisAnswer: (c)
Q.27
A non planer loop of conducting wire carrying a current I is placed as shown in figure. Each of the straight sections of the loop is of length 2a. the magnetic field due to this loop at the point P(a, 0, a) point in the direction [ IIT 2001]
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a)
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b)
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c)
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d)
Explanation
If we take individual length for the purpose of calculating magnetic filed in a 3-dimensional figure then it will be difficult Here we will consider two loops ABEF and BCDE. There is no conductor along BE we will assume that there are two conductors one carry current along BE while other carry conductor along EB, hence magnetic filed gets cancelled Loop ABEF :Loop is in X-Y plane , magnetic filed will be in Z direction at point (a,0, a) LOOP BCDE:Loop is in X-Y plane , magnetic filed will be along X direction at point(a,0, a) Magnitude of magnetic filed due to both loop at point will be same, hence resultant will be along position vector of point at (a,0,2) ∴ unit vector will be (1/√2) (i + j)Answer: (c)
Q.28
Two particles A and b of masses ma and mb respectively and having the same charge are moving in plane. A uniform magnetic field exists perpendicular to this plane. the speeds of the particles are va and vb respectively and the trajectories are as shown in figure. Then [ IIT 2001]
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a)mava < mbvb
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b)mava > mbvb
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c)ma < mb and va < vb
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d)ma=mb and va=vb
Explanation
When a charged particle is moving at right angle to the magnetic filed then force act on it which behaves as a centripetal force and moves the particle in circular motion Answer:(b)
Q.29
A coil having N turns is wound tightly in the form of spiral with inner and outer radii a and b respectively. When a current I passes through the coil, the magnetic at center is [ IIT 2001]
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a) µNI / b
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b) 2µNI / a
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c)
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d)
Explanation
Let us consider a thickness dx of wire. Let it be a distance x from the centre O. Number of turns per unit length=N /(b-a) ∴ Number of turns in thickness dx=[ N /(b-a)]dx Small amount of magnetic field is produced at O due to thickness dx of the wire . Answer: (c)
Q.30
A particle of mass m and charge q moves with a constant velocity V along the positive x-direction. It enters a region containing a uniform magnetic field B directed along the negative z-direction, extending from x=a to x=b. the minimum value of v required so that the particle can just enter the region x > b is [ IIT 2002]
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a)qbB/m
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b) [q(b-a)B] / m
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c)qaB / m
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d)[q(b+a)B] / 2m
Explanation
Particle has to pass through the magnetic filed region whose width is ( b-a).When charged particle reenters a magnetic filed it will follow a circular path of radius Raccording to required R ≥ (b-a) thus Answer: (b)Q65) A particle of mass m and charge q moves with a constant velocity V along the positive x-direction. It enters a region containing a uniform magnetic field B directed along the negative z-direction, extending from x=a to x=b. the minimum value of v required so that the particle can just enter the region x > b is [ IIT 2002]a)qbB/m b) [q(b-a)B] / m c)qaB / m d)[q(b+a)B] / 2m SOLUTION Particle has to pass through the magnetic filed region whose width is ( b-a).When charged particle reenters a magnetic filed it will follow a circular path of radius Raccording to required R ≥ (b-a) thus Answer: (b)
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