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. 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
- a)(q(b+a)B)/2m
- b)(qaB)/m
- c)(q(b-a)B)/m
- d)(qbB)/m
Correct answer is option 'C'. Can you explain this answer?
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.A particle of mass m and charge q moves with a constant velocity v al...
To determine the minimum value of v required for the particle to enter the region x
The magnetic force on a charged particle moving in a magnetic field is given by the equation:
F = qvBsinθ
Where F is the magnetic force, q is the charge of the particle, v is its velocity, B is the magnetic field, and θ is the angle between the velocity vector and the magnetic field vector.
Since the particle is moving along the positive x-direction and the magnetic field is along the negative z-direction, the angle θ between the velocity and the magnetic field is 90 degrees.
sin(90) = 1
Therefore, the equation for the magnetic force simplifies to:
F = qvB
In order for the particle to enter the region x
Since the particle is moving with a constant velocity, the net force acting on it must be zero. Therefore, the magnetic force must balance out with an equal and opposite force.
The only force that can balance out the magnetic force is the electric force:
Fe = qE
Where Fe is the electric force and E is the electric field.
Since the particle is moving with a constant velocity, the electric field must be directed along the positive x-direction, opposing the particle's motion.
In order for the particle to enter the region x
Therefore, we can write the equation for the electric force as:
Fe = qE
Since the electric force must balance out the magnetic force, we can equate the two forces and solve for the electric field:
qvB = qE
E = vB
The minimum value of v required for the particle to enter the region x
v = E/B
Since E is the electric field and B is the magnetic field, the minimum value of v required is equal to the ratio of the electric field to the magnetic field.
The magnetic force on a charged particle moving in a magnetic field is given by the equation:
F = qvBsinθ
Where F is the magnetic force, q is the charge of the particle, v is its velocity, B is the magnetic field, and θ is the angle between the velocity vector and the magnetic field vector.
Since the particle is moving along the positive x-direction and the magnetic field is along the negative z-direction, the angle θ between the velocity and the magnetic field is 90 degrees.
sin(90) = 1
Therefore, the equation for the magnetic force simplifies to:
F = qvB
In order for the particle to enter the region x
Since the particle is moving with a constant velocity, the net force acting on it must be zero. Therefore, the magnetic force must balance out with an equal and opposite force.
The only force that can balance out the magnetic force is the electric force:
Fe = qE
Where Fe is the electric force and E is the electric field.
Since the particle is moving with a constant velocity, the electric field must be directed along the positive x-direction, opposing the particle's motion.
In order for the particle to enter the region x
Therefore, we can write the equation for the electric force as:
Fe = qE
Since the electric force must balance out the magnetic force, we can equate the two forces and solve for the electric field:
qvB = qE
E = vB
The minimum value of v required for the particle to enter the region x
v = E/B
Since E is the electric field and B is the magnetic field, the minimum value of v required is equal to the ratio of the electric field to the magnetic field.
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.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 isa)(q(b+a)B)/2mb)(qaB)/mc)(q(b-a)B)/md)(qbB)/mCorrect answer is option 'C'. Can you explain this answer?
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.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 isa)(q(b+a)B)/2mb)(qaB)/mc)(q(b-a)B)/md)(qbB)/mCorrect answer is option 'C'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about .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 isa)(q(b+a)B)/2mb)(qaB)/mc)(q(b-a)B)/md)(qbB)/mCorrect answer is option 'C'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for .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 isa)(q(b+a)B)/2mb)(qaB)/mc)(q(b-a)B)/md)(qbB)/mCorrect answer is option 'C'. Can you explain this answer?.
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Solutions for .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 isa)(q(b+a)B)/2mb)(qaB)/mc)(q(b-a)B)/md)(qbB)/mCorrect answer is option 'C'. Can you explain this answer? in English & in Hindi are available as part of our courses for JEE.
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