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A charged particle with q/m =2 is projected with velocity 5m/s i in a region of uniform magnetic field 4i 3j exist if maximum possible displacement along z axis is 3/k meters then k is ?
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A charged particle with q/m =2 is projected with velocity 5m/s i in a ...
Explanation of the Problem


The problem deals with the motion of a charged particle in a uniform magnetic field. The charge to mass ratio is given as 2, and the velocity of the particle is 5 m/s in the x-direction. The magnetic field is given as 4i + 3j. We need to find the maximum possible displacement along the z-axis, given as 3/k meters.

Solution


Force on the Charged Particle


The force experienced by a charged particle in a magnetic field is given by the Lorentz force equation:

F = q (v x B)

where F is the force, q is the charge of the particle, v is the velocity of the particle, and B is the magnetic field.

In this problem, the charge to mass ratio is given as 2, and the velocity of the particle is 5 m/s in the x-direction. The magnetic field is given as 4i + 3j.

So, the force experienced by the particle is given by:

F = 2 (5i x (4i + 3j))

F = 2 (15k)

F = 30k

Therefore, the force experienced by the particle is 30k.

Motion of the Charged Particle


The motion of a charged particle in a magnetic field can be described by the equation:

F = ma

where F is the force on the particle, m is the mass of the particle, and a is the acceleration of the particle.

In this problem, the mass of the particle is not given. However, we can use the charge to mass ratio to find the mass of the particle.

q/m = 2

q = 2m

Therefore, m = q/2

m = (2/2)

m = 1

So, the mass of the particle is 1.

Therefore, the equation of motion of the particle is given by:

F = ma

30k = 1a

a = 30k

The acceleration of the particle is 30k.

Maximum Displacement along Z-Axis


The maximum displacement along the z-axis can be found using the following equation:

z = (1/2) (a z t^2)

where a z is the acceleration in the z-direction, and t is the time of flight.

In this problem, the particle is projected with velocity 5 m/s in the x-direction. Therefore, the velocity of the particle in the z-direction is zero.

So, the acceleration in the z-direction is zero.

Therefore, the maximum displacement along the z-axis is zero.

However, the problem statement gives the maximum possible displacement along the z-axis as 3/k meters.

Therefore, k must be infinity.

Conclusion


In conclusion, we have found that the maximum displacement along the z-axis is zero, and k must be infinity. The problem deals with the motion of a charged particle in a uniform magnetic field, and we have used the Lorentz force equation and the equation of motion to solve the problem.
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A charged particle with q/m =2 is projected with velocity 5m/s i in a region of uniform magnetic field 4i 3j exist if maximum possible displacement along z axis is 3/k meters then k is ?
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A charged particle with q/m =2 is projected with velocity 5m/s i in a region of uniform magnetic field 4i 3j exist if maximum possible displacement along z axis is 3/k meters then k is ? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about A charged particle with q/m =2 is projected with velocity 5m/s i in a region of uniform magnetic field 4i 3j exist if maximum possible displacement along z axis is 3/k meters then k is ? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A charged particle with q/m =2 is projected with velocity 5m/s i in a region of uniform magnetic field 4i 3j exist if maximum possible displacement along z axis is 3/k meters then k is ?.
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