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Physics NEET MCQ
Magnetic Effects Of Current Mcq
Quiz 3
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Q.1
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.2
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.3
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.4
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.5
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.6
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.7
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)
Q.8
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.9
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.10
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.11
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.12
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.13
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.14
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.15
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.16
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.17
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.18
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.19
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.20
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.21
A particle of charge -10×10-18 coulomb moving with velocity 10ms-1 along the x axis enters a region where a magnetic field induction B is along the y-axis, and an electric filed of magnitude 104 V/m is along the negative z-axis. If the charged particle continues moving along the x-axis, the magnitude of b is [ AIEEE 2003]
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a) 103 Wb/m2
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b) 105 Wb/m2
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c) 1016 Wb/m2
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d) 10-3 Wb/m2
Explanation
The situation is as shown in figure FB = Force due to magnetic field FE = Force due to electric field It is given that the charged particle remains moving along X-axis ( i.e. un deviated) Therefore FB = FE qvB = qE B = E/v = 104/10 = 103 weber / m2 Answer: (a)
Q.22
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.23
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.24
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.25
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.26
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.27
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.28
A long solenoid has 200 turns per cm and carries a current i. The magnetic field at its centre is 6.28×10-2 Weber/mAnother long solenoid has 100turns per cm and it carries a current i/The value of the magnetic field at its centre is [ AIEE 2006]
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a) 1.05×10-2 Weber/ m2
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b) 1.05×10-5 Weber/ m2
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c) 1.05×10-3 Weber/ m2
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d) 1.05×10-4 Weber/ m2
Explanation
Answer: (a)
Q.29
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.30
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)
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