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Physics NEET MCQ
Quiz 4
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Q.1
Current of 10 amp. and 2 amp. are passed through two parallel wire A and B respectively in opposite directions. If the wire A is infinitely long and the length of the wire B is 2 metres, the force on the conductor B, which is situated at 10 cm distance from A will be [ CPMT 1988]
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c)8π×10-7 N
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d)4π×10-7 N
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a) 8×10⁻⁵ N
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b) 4×10⁻⁵ N
Explanation
Force between wires is given by Answer: (a)
Q.2
A horizontal rod of mass 10g and length 10cm is placed on smooth inclined plane making an angle 60° with horizontal, with the length of the rod parallel to the edge of inclined plane. A uniform magnetic field of induction B is applied vertically downwards. If the current through the rod is 1.73 amp. the value of B for which the rod remains stationary on the inclined plane is
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a) 1.73 tesla
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b) 1 /1.73 tesla
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c)1 tesla
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d)0.5 tesla
Explanation
Current is coming out of the paper. By Fleming's left hand rule force on the conductor is BIL as angle between direction of current and magnetic field is 90°, component of force parallel to inclined plane is BIlcosθ From figure BIlcosθ=mgsinθ Answer:(a)
Q.3
A small coil of N turns has an effective area A and carries a current I. It is suspended in a horizontal magnetic field B such that its plane is perpendicular to B. The work done in rotating it by 180° about the vertical axis is [ MPPMT 1994]
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a) NAIB
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b) 2NIAB
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c) 2πNAIB
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d) 0
Explanation
W=MB( 1- cos180)=2MB For current carrying coil M=NIA ∴ W=2MBI Answer: (b)
Q.4
A rectangular loop carrying current is placed near a long straight fixed wire carrying strong current such that long side are parallel to wire. If the current in the nearer long side of loop is parallel to current in the wire. Then the loop [ MPPMT 1999]
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a)experiences no force
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b) experiences a force towards the wire
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c)experiences force away from wire
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d)experiences a torque but no force
Explanation
The force on the nearer arm of the loop is towards the wire (left) because of attraction ( current in same direction) while the force on the farther arm is away from the loop ( right) but since F∝ (1/r), the force on the nearer arm is greater and so the loop shifts towards the wireAnswer: (b)
Q.5
The unit of electric current "ampere" is the current which when flowing through each of two parallel wires spaced 1 m apart in vacuum and of infinite length will give rise to a force between them equal to [ MPPMT 1999]
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a) 1 N/m
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d)4π×10-7 N/m
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b) 2×10⁻⁷ N/m
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c)1×10⁻² N/m
Explanation
Answer: (b)
Q.6
An electron is shot in steady electric and magnetic filed such that its velocity v,. Electric field and magnetic fields are perpendicular to each other and direction of electron. There strengths are such that they cancels each other effect . Magnitude of E is 1 volt/m and B is 2 tesla. Then velocity of electron is [ BIT 1988]
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a) 50 m/s
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b) 2 m/s
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c)0.5 cm/s
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d)200 m/s
Explanation
Since electron goes un deflected E=Bv E=1V/cm=100 V/mv=E/ B=100/2=50 m/s Answer:(a)
Q.7
Following figure (1) and (2) represent lines of force. Which of the following is correct statement? [ MPPET 1995]
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a) Figure (1) represents magnetic lines of force
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b) Figure (2) represents magnetic lines of force
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c) Figure (1) represents electric lines of force
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d) Both figure (1) and figure(2) represent magnetic lines of force
Explanation
Magnetic field lines can form a closed circular loop Answer: (a)
Q.8
In the given diagram two long parallel wires carry equal currents in opposite directions. Point O is situated midway between the wires and the xy-plane contains two wires and the positive z-axis comes normally out of the plane of paper. the magnetic field B at O is non-zero along : [ SCRA 1994]
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a)X, Y and Z axes
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b) X-axis
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c)Y-axis
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d)Z-axis
Explanation
Using right hand thumb rule we find direction along Z axis which is perpendicular to plane of paper. Answer:(d)
Q.9
A conducting wire is moving towards right in a magnetic field B. The direction of induced current in the wire is shown in the figure. the direction of magnetic field will be [ MPPMT 1995]
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a) in the plane of paper pointing towards right
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b) in the plane of paper pointing towards left
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c) perpendicular to the plane of paper an downwards
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d) perpendicular to the plane of paper and upwards
Explanation
y applying Fleming's right hand rule . direction of B will be perpendicular to the plane and going downwards Answer: (c)
Q.10
A positively charged particle moving with velocity V enters a region of space having a constant magnetic induction B. The particle will experience the largest deflecting force when the angle between the vector V and B is [ MPPMT 1998]
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a)0°
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b) 45°
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c)90°
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d)180°
Explanation
Election caused due to magnetic force on moving charged particle F=q(V×B) thus maximum deflection is at angle 90°Answer: (c)
Q.11
A proton and an electron both moving with same velocity v enters into a region of magnetic field directed perpendicular to the velocity of the particles. They will now move in circular orbits such that [ PMT 1995]
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a) their time period will be same
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b) the time period for proton will be higher
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c)the time period for electron will be higher
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d)their orbital radii will be same
Explanation
Time period T=2πm / qB charge on proton are equal in magnitude Thus T ∝ m mass of proton is more than mass of electron thus time period for proton will be moreAnswer: (b)
Q.12
A particle moving in a magnetic field has increase in its velocity, then the radius of circle [ BHU 1998]
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a) decreases
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b) increases
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c)remains the same
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d)becomes half
Explanation
radius r=p/qB Thus r ∝ Answer:(b)
Q.13
A moving coil galvanometer has N number of turns in a coil of effective it carries a current I. the magnetic field B is radial. The torque acting on the coil is [ MPPMT 1994]
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a) NA2B2I
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b) NABI2
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c) N2ABI
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d) NABI
Explanation
Answer: (d)
Q.14
Two particles X and Y having equal charge after being accelerated through the same potential difference, enters a region of uniform magnetic field and describe circular paths of radii R1 and R2 respectively. The ratio of the mass of X and that of Y is [ CBSE 1995]
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a) (R₁ / R₂)1/2
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b) R₂ / R₁
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c)(R₁ / R₂)²
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d)R₁ / R₂
Explanation
Energy of charged particles will be same momentum p=√2EM and radius r=p / qB From above r ∝ √m Thus R1 / R2=√ ( m1 / m2) Thus m1 / m2=[R1 / R2]2Answer: (c)
Q.15
An infinitely long straight conductor is bent into shape as shown in figure. It carries a current i amp. and the radius of circular loop is r, then magnetic field at centre of the circular loop is [ MPPMT 1999]
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a) 0
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b) ∞
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c)
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d)
Explanation
Direction of magnetic induction at O is perpendicular to paper outward . While magnetic induction at O due to infinite long wire is perpendicular to paper inward, they are opposite to each other Magnetic filed at the centre of loop Magnetic field at the point O due to infinitely long wire Resultant magnetic filed B Answer:(d)
Q.16
An electron and proton enters a region of uniform magnetic field with the same kinetic energy. They describe circular paths of radius re and rp respectively. Then [ CPMT 1999]
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a) re=rp
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b) re < rp
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c) re > rp
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d) re may be less than or greater than rp depending on the direction of magnetic field
Explanation
given angle between direction of velocity and magnetic field 90° We now that momentum p=√(2Em) Here E is kinetic energy and m is the mass of particle and p=qrB from above equations qrB=√(2Em) Thus r ∝ √m mass of proton is grater than mass of electron ∴ rp > re Answer: (b)
Q.17
A proton moving with constant velocity passes through a region of space without changing in its velocity. If E and B represents electric and magnetic filed respectively, this region of space may not have [ AMU 1995]
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a)E=0, B=0
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b) E=0, B ≠ 0
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c)E ≠ 0, B=0
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d)E ≠ 0, B ≠ 0
Explanation
When charge passes through magnetic and electric field with out changing speed is possible if i) no electric and magnetic field presentii) Electric field is zeroiii) electric field and magnetic fields are perpendicular to each other such that they cancel out each other effect OPTIONS ARE FOR MAY NOT HAVE thus option c is the best option Answer: (c)
Q.18
A certain wire of length L carries a current I. It is bent to form a circle of one turn. The magnetic field at the centre of the loop is B. The same wire now is made to form a circular loop of two turns. The magnetic field now at centre is [ MPPMT 1999]
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a) 2B
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b) 4B
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c)B/2
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d)B/4
Explanation
When same length wire is bent to give n turns . Then B'=n2B Here B is magnetic field due to single turn and n are the number of turn Thus B'=(2)2B=4BAnswer: (b)
Q.19
A very long solenoid has 800 turns per metre length of solenoid. A current of 1.6 amp. flows through it. Then the magnetic induction at the end of the solenoid on its axis is [ MPPMT 1999]
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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
From formula Answer:(b)
Q.20
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.21
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.22
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.23
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.24
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.25
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.26
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.27
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.28
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.29
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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d)2×1019 µo
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c)2×10⁻¹⁹ 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.30
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)
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