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Physics NEET MCQ
Quiz 3
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Q.1
A long straight wire along the Z-direction carries a current I in the negative X-direction. The magnetic vector field B at a point having coordinates (x, y) in the Z=0 plane is [ IIT 2002]
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a)
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b)
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c)
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d)
Explanation
The wire carries a current I in the negative Z direction. We have to consider vector field B at (x,y) in the z=0 plane Magnetic field B is perpendicular to OP∴ B=Bsinθ i - Bcosθj sinθ=y/r and cosθ=x/r and B=(µoI) / (2πr)Answer: (a)
Q.2
For a positively charged particle moving in x-y plane initially along the x-axis, there is a sudden change in its path due to the presence of electric and/or magnetic filed beyond P. The curve path is shown in the x-y plane and is found to be non-circular which one of the following combinations is possible? [ IIT 2003]
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a) E=0, B=bi + ck
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b) E=ai, B=ck + ai
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c)E=0, B=cj + bk
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d)E=ai, B=ck + bj
Explanation
The velocity at P is in the X-direction (given)Let V=niAfter P, the positive charged particle gets deflected in the x-y plane towards -y direction and path is non-circularNow F=q(v×B) From option bF=q[ni×(ck + ai)] F=q[nci×k + mai× i)] F=ncq(-j)Since in option b electric field is along x axis will accelerate the particle in positive x-direction, where as magnetic field will move particle in negative y-direction as a a result of two forces path will be non-circular . Option 'b' correct Answer:(b)
Q.3
A conducting loop carrying a current i is placed in a uniform magnetic field pointing into the plane of the paper as shown in figure. The loop will have a tendency to [ IIT 2003]
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a) contract
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b) expand
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c) move towards +x axis
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d) move towards -x axis
Explanation
Use Fleming's left hand rule. We find that a force is acting in the radial outward direction throughout the circumference of the conducting loop Answer: (b)
Q.4
A current carrying loop is placed in a uniform magnetic field in four different orientations, I, II, III and IV arrange them in the decreasing order of Potential Energy [ IIT 2003]
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a)I > III > II > IV
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b)I > II > III > IV
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c)I > IV > II > III
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d)III > IV > I > II
Explanation
We know that U=-M . BU=-MBcosθ In case I, θ=180°, U=+MBCase II, θ=90°, U=0Case III, θ=acute, U=positive less than MBCase IV, θ=obtuse, U=negative∴ I > III > II > IVAnswer: (a)
Q.5
An electron traveling with a speed of u along the positive x axis enters into the region of magnetic field where B=-Bok ( x >0). It comes out of the region with speed v then [ IIT 2004]
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a) v=u at Y>0
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b) v=u at y < 0
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c)V > u at y > 0
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d)v > u at y < 0
Explanation
the force acting on electron will be perpendicular to the direction of velocity till the electron remains in the magnetic field. So the electron will follow the path as shown in figure Answer: (b)
Q.6
A charged particle is moving with velocity v in a uniform magnetic field B. The magnetic force acting on it will be maximum when .....
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a) v and B are in the same direction
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b) v and B are in the opposite directions
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c) v and B are mutually perpendicular
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d) v and B makes an angle of 45° with each other
Explanation
From formula F=qvbsinθ Answer: (c)
Q.7
When equal current pass through two very long and straight parallel wires in mutually opposite direction then they ....
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a)attract each other
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b) repel each other
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c)lean towards each other
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d)neither repel nor attract
Explanation
Answer: (b)
Q.8
A magnetic field B=Boj, exists in the region a < x < 2a and B=-Boj, in the region 2a < x < 3a, where Bo is positive constant. A positive point charge moving with velocity v=voi, where vo is a positive constant, enters the magnetic field at x=a. The trajectory of the charge in the region can be like [ IIT 2007]
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a)
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b)
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c)
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d)
Explanation
We can find direction using vector form of B and v F=q(v×B) For a < x < 2a : F=q(voi × Boj) Thus direction of particle will be along z direction For 2a < x < 3a : F=q(vok × -Boj)Thus direction will be along positive x axis Answer:(a)
Q.9
If in circular coil A of radius R, current I is flowing and in another coil B of radius 2R a current 2I is flowing, then the ratio of the magnetic fields BA and BB produced by them will be [ AIEEE 2002]
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a) 1
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b) 2
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c) 1/2
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d) 4
Explanation
From the formula for magnetic field produced by a current carrying circular loop at its centre is B ∝ I / r Thus BA ∝ I/R BA ∝ 2I/2R BA / BB=1 Answer: (a)
Q.10
If an electron and proton having same momentum enter perpendicular to magnetic field, then [ AIEEE 2002]
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a) curved path of electron and proton will be same (ignore the sense of revolution)
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b) they will be move un deflected
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c) curve path of electron is more curve than that of proton
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d) path of proton is more curved
Explanation
When a charged particle enters perpendicular to a magnetic field, then it moves in a circular path of radius r = p / (qB) Here p is momentum, q charge and B is magnetic field Since all the quantities fro electron and proton are same radius of both will be same Answer: (a)
Q.11
The time period of a charged particle undergoing a circular motion in a uniform magnetic field is independent of its [ AIEEE 2002]
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a) speed
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b) mass
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c) charge
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d) magnetic induction
Explanation
Time period of a charged particle moving in a magnetic filed (B) is T = 2πm / (qB) The time period does not depend on the speed of the particle Answer: (a)
Q.12
A particle of mass M and charge Q moving with velocity v describe a circular path of radius R when subjected to a uniform transverse magnetic filed of induction B. The work done by the field when the particle complete one full circle is [ AIEEE 2003]
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a)
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b) zero
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c) BQ2πR
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d) BQv2πR
Explanation
Charge particle follows a circular path under the effect of magnetic force thus force is perpendicular to displacement work done is zero Answer:(b)
Q.13
Wire 1 and 2 carrying currents i1 and i2 respectively are inclined at an angle of θ to each other. What is the force on a small element dl of wire 2 at a distance of r from wire 1 ( as shown in figure) due to the magnetic field of wire 1? [ AIEEE 2002]
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a)
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b)
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c)
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d)
Explanation
Magnetic field due to current in wire 1 at point P distant r from the wire is direction of magnetic field is perpendicular to the plane of paper, inward The force exerted due to this magnetic field on current element i2dl is dF = i2dlBsin90 Answer: (c)
Q.14
A current i ampere flows along an infinitely straight thin walled tube, then the magnetic induction at any point inside the tube is [ AIEEE 2004]
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a) µo2i / (4πr) tesla
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b) zero
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c) infinite
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d) 2i/r tesla
Explanation
Using Ampere's circuital law at a distance r from the axis of tube r< radius of tube No current is enclosed by the loop thus magnetic induction is zero Answer: (b)
Q.15
A long wire carrie a steady current. It is bent into a circle of one turn and the magnetic field at centre of the coil is B. It is then bent into a circular loop of n turns. the magnetic field at the centre of the coil will be [ AIEEE 2004]
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a) 2nB
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b) n2B
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c) nb
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d) 2n2B
Explanation
Magnetic field at the centre of a circular coil of n turns is given by Thus B ∝ N/R if current is constant Given: n×(2πr) = 2πR R = nr or r = R/n B' ∝ n / ( R/n) B' ∝ n2 / R Thus B' /B = n2B Answer:(b)
Q.16
The magnetic field due to a current carrying circular loop of radius 3cm at a point on the axis at a distance of 4 cm from the centre is 54µT. What will be its value at the centre of the loop? [ AIEEE 2004]
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a) 125µT
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b) 150 µT
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c) 250µT
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d) 75µT
Explanation
The magnetic field at a pont on the axis of circular loop at a distance x from centre is here a is radius of loop Magnetic field at the centre of loop B' By taking ratio of B and B' we get Answer: (c)
Q.17
Two concentric coils each of radius equal to 2π cm are placed at right angles to each other. 3 amp and 4 amp. are the currents flowing in each coil respectively. The magnetic induction is Weber / m2 M at the centre of the coil will be ( µo=4π× 10-7 Wb/A.m) [AIEEE 2005]
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a)10-5
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b) 12×10⁻⁵
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c)7×10⁻⁵
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d)5×10⁻⁵
Explanation
Magnetic field due to circular coilsMagnetic filed produced by coils is mutually perpendicular thusAnswer: (d)
Q.18
A charged particle of mass m and charge q travels on circular path of radius r that is perpendicular to a magnetic field B. The time taken by the particle to complete one revolution is [ AIEEE 2005]
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a) (2πq2B) / m
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b) (2πmq)/B
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c)2πm / qB
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d)2πqB / m
Explanation
Equating magnetic force to centripetal forceAnswer: (c)
Q.19
A uniform electric field and a uniform magnetic field are acting along the same direction in a certain region. If an electron is projected along the direction of the field with a certain velocity the [ AIEEE 2005]
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a) its velocity will increase
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b) its velocity will decrease
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c)it will turn towards left of direction of motion
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d)it will turn towards right of direction of motion
Explanation
Direction of motion of electron and direction of electric field is same. Electron will experience force in opposite to direction of its motion , velocity will decrease Answer:(b)
Q.20
If a charged particle is moving through a uniform magnetic field, then its .....
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a) momentum changes but kinetic energy does not change
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b) both momentum and kinetic energy chang
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c)momentum and kinetic energy do not change
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d)kinetic energy changes but momentum does not change
Explanation
Answer: (a)
Q.21
If the speed of the particle moving through a magnetic field is increased, then the radius of curvature of the trajectory will ....
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a) decrease
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b) increase
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c)not change
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d)become half
Explanation
we know that r=mv/ Be ∴ r ∝ v Answer:(b)
Q.22
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.23
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.24
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.25
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.26
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.27
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.28
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.29
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.30
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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c) µo i / 2πb
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d) µo i / 4πb
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a) µ o i² / b²
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b) µo i² / (2π b²)
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)
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