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Physics NEET MCQ
Magnetic Effects Of Current Mcq
Quiz 4
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Q.1
Weber/ m2=......
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a) volt
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b) henry
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c) tesla
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d) all the three
Explanation
Answer: (c)
Q.2
A long straight wire of radius a carries a steady current i. The current is uniformly distributed across its cross-section. The ratio of the magnetic field at a/2 and 2a is [ AIEEE 2007]
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a)1/2
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b) 1/4
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c)4
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d)1
Explanation
Current is uniformly distributed thus current per unit area=i / πa2Thus current enclosed by the wire of radius r1 From Ampere's la magnetic field B1=(µo×Total current)/ pathFor r=2a current enclosed=i thus Thus ratio B1 / B2=1 Answer: (d)
Q.3
A current I flows along the length of an infinitely long straight, thin walled pipe. then [ AIEEE 2007]
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a) the magnetic field at all points inside the pipe is the same, but not zero
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b) the magnetic field is zero only on the axis of the pipe
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c)the magnetic field is different at different point inside the pipe
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d)the magnetic field at any point inside the pipe is zero
Explanation
There no current inside the pipe thus magnetic field is zero, according to Ampere's circuital lawAnswer: (d)
Q.4
A charged particle with charge q enters a region of constant, uniform and mutually orthogonal field B and E with a velocity v perpendicular to both E and B, and comes out without any change in magnitude or direction of v. Then [ AIEEE 2007]
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a) v=(B×E) / E2
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b) v=(E×B) / B2
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c) v=(B×E) / B2
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d) v=(E×B) / E2
Explanation
Here, E and B are perpendicular to each other and the velocity v does not change, qE=qvBv=E/B Also Answer:(b)
Q.5
A charged particle moves through a magnetic field perpendicular to its direction. Then [ AIEEE 2007]
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a) kinetic energy changes but momentum is constant
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b) the momentum changes but kinetic energy is constant
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c) both momentum and kinetic energy of the particle are not constant
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d) both momentum and kinetic energy of the particle are constant
Explanation
When charged particle flows a circular path in magnetic field, direction of velocity changes thus momentum charges as it is also a vector. But magnitude of velocity remains constant thus kinetic energy do not change Answer: (b)
Q.6
Two identical conducting wires AOB and COD are placed at right angels each other. The wire AOB carries an electric current I1 and COD carries a current IThe magnetic field on a point lying at a distance d from O, in a direction perpendicular to the plane of the wire AOB and COD will be given by [ AIEEE 2007]
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a)
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b)
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c)
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d)
Explanation
Clearly the magnetic field at point P, equidistant from AOB and COD will have directions perpendicular to each other, as they are placed normal to each otherResultant field B Answer: (c)
Q.7
A horizontal overhead power line is at height of 4m from the ground and carries a current of 100A from east to west. The magnetic field directly below it on the ground is ( µo=4π×10-7 TMA-1) [ AIEEE 2008]
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a) 2.5×10-7 T southward
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b) 5×10-6 T northward
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c)5×10-6 T southward
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d)2.5×10-7 T northward
Explanation
The magnetic field is According to right hand palm rule, the magnetic field directed towards southAnswer: (c)
Q.8
Relative permitivity and permeability of a material εr and µr respectively. Which of the following values of these quantities are allowed for a diamagnetic material? [ AIEEE 2008]
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a)εr=0.5, µr=1.5
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b) εr=1.5, µr=0.5
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c)εr=0.5, µr=0.5
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d)εr=1.5, µr=1.5
Explanation
For a diamagnetic material, the value of µr is less than one. For any material, the value of εr is always greater than 1 Answer:(b)
Q.9
Two thin long wires parallel to each other separated by a distance b are carrying a current i amp. each. The magnitude of the force per unit length exerted by one wire on the other is [ Raj. PMT 1997]
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a) µ o i2 / b2
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b) µo i2 / (2π b2)
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c) µo i / 2πb
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d) µo i / 4πb
Explanation
Magnetic field produced by first long wire B=µo i / 2πb This magnetic filed is perpendicular to second wire Thus force on second wire=Bil here l is the length of second wire Thus F=(µo i / 2πb) il F/l=µo i2 / (2π b2) Answer: (b)
Q.10
Two straight parallel wires both carrying 1 amp current in the same direction attracts each other with a force of 1×10-3 N. If both the currents are doubled, the force of attraction will be [ MPPMT 1994]
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a)1×10-3 N
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b) 2×10-3 N
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c)4×10-3 N
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d)0.25×10-3 N
Explanation
We know that Force per unit length F ∝ i1 I2 if distance is constantThus by making current double in each wire force will be four times of initialThus force will be 4×10-3 NAnswer: (c)
Q.11
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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a) 8×10-5 N
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b) 4×10-5 N
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c)8π×10-7 N
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d)4π×10-7 N
Explanation
Force between wires is given by Answer: (a)
Q.12
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.13
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.14
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.15
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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b) 2×10-7 N/m
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c)1×10-2 N/m
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d)4π×10-7 N/m
Explanation
Answer: (b)
Q.16
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.17
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.18
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.19
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.20
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.21
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.22
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.23
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.24
A particle with 10-11 coulomb charge and 10-7 kg mass is moving with velocity of 108 m/s along the y-axis. A uniform static magnetic field B=0.5 Tesla is acting along x-direction. The force on the particle is [ MPPMT 1997]
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a)5×10-11 N along i
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b) 5×103 N along k
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c)5×10-11 N along -j
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d)5×10-4 N along -k
Explanation
F=q(v×B) F=10-11(108j×0.5i) F=5×10-4(-k)Answer: (d)
Q.25
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.26
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.27
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.28
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.29
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.30
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)
0 h : 0 m : 1 s
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