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A particle of charge -16*10^-18C moving with a velocity of 10m/s along the x-axis enters a region where the magnetic field is along the -ve z-axis. If the charged particle continues moving along the x-axis, the magnitude of magnetic field is a. 10^3 Wb/m^2 b. 10^5Wb/m^2 c. 10^16 Wb/m^2 d. 10^-3 Wb/m^2. pls explain the solution?
Most Upvoted Answer
A particle of charge -16*10^-18C moving with a velocity of 10m/s along...
The Lorentz Force
To solve this problem, we need to use the Lorentz force formula, which describes the force experienced by a charged particle moving in a magnetic field. The Lorentz force (F) is given by the equation:
F = q(v x B)
where q is the charge of the particle, v is its velocity, and B is the magnetic field vector.
Given Information
In this case, we are given:
- Charge of the particle (q) = -16 * 10^-18 C
- Velocity of the particle (v) = 10 m/s along the x-axis
- Magnetic field (B) is along the -ve z-axis
Applying the Lorentz Force Formula
Since the particle continues to move along the x-axis, the magnetic force must be zero in that direction. Therefore, we can conclude that the magnetic field (B) does not have any x-component. This means that Bx = 0.
Now, let's calculate the magnetic force in the y-direction (Fy) using the Lorentz force formula:
Fy = q(vyBz)
Since the velocity is along the x-axis, vy = 0. Therefore, Fy = 0.
Similarly, calculating the magnetic force in the z-direction (Fz):
Fz = q(vzBz)
Since the velocity is along the x-axis, vz = 0. Therefore, Fz = 0.
Conclusion
From the above calculations, we can see that the Lorentz force in both the y and z directions is zero. This implies that the particle does not experience any magnetic force and continues moving along the x-axis unaffected by the magnetic field.
Hence, the magnitude of the magnetic field (B) does not affect the motion of the particle in this case. Therefore, the correct answer is d) 10^-3 Wb/m^2.
To solve this problem, we need to use the Lorentz force formula, which describes the force experienced by a charged particle moving in a magnetic field. The Lorentz force (F) is given by the equation:
F = q(v x B)
where q is the charge of the particle, v is its velocity, and B is the magnetic field vector.
Given Information
In this case, we are given:
- Charge of the particle (q) = -16 * 10^-18 C
- Velocity of the particle (v) = 10 m/s along the x-axis
- Magnetic field (B) is along the -ve z-axis
Applying the Lorentz Force Formula
Since the particle continues to move along the x-axis, the magnetic force must be zero in that direction. Therefore, we can conclude that the magnetic field (B) does not have any x-component. This means that Bx = 0.
Now, let's calculate the magnetic force in the y-direction (Fy) using the Lorentz force formula:
Fy = q(vyBz)
Since the velocity is along the x-axis, vy = 0. Therefore, Fy = 0.
Similarly, calculating the magnetic force in the z-direction (Fz):
Fz = q(vzBz)
Since the velocity is along the x-axis, vz = 0. Therefore, Fz = 0.
Conclusion
From the above calculations, we can see that the Lorentz force in both the y and z directions is zero. This implies that the particle does not experience any magnetic force and continues moving along the x-axis unaffected by the magnetic field.
Hence, the magnitude of the magnetic field (B) does not affect the motion of the particle in this case. Therefore, the correct answer is d) 10^-3 Wb/m^2.
Community Answer
A particle of charge -16*10^-18C moving with a velocity of 10m/s along...
Is the value of electric field also needed?
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A particle of charge -16*10^-18C moving with a velocity of 10m/s along the x-axis enters a region where the magnetic field is along the -ve z-axis. If the charged particle continues moving along the x-axis, the magnitude of magnetic field is a. 10^3 Wb/m^2 b. 10^5Wb/m^2 c. 10^16 Wb/m^2 d. 10^-3 Wb/m^2. pls explain the solution?
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A particle of charge -16*10^-18C moving with a velocity of 10m/s along the x-axis enters a region where the magnetic field is along the -ve z-axis. If the charged particle continues moving along the x-axis, the magnitude of magnetic field is a. 10^3 Wb/m^2 b. 10^5Wb/m^2 c. 10^16 Wb/m^2 d. 10^-3 Wb/m^2. pls explain the solution? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about A particle of charge -16*10^-18C moving with a velocity of 10m/s along the x-axis enters a region where the magnetic field is along the -ve z-axis. If the charged particle continues moving along the x-axis, the magnitude of magnetic field is a. 10^3 Wb/m^2 b. 10^5Wb/m^2 c. 10^16 Wb/m^2 d. 10^-3 Wb/m^2. pls explain the solution? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A particle of charge -16*10^-18C moving with a velocity of 10m/s along the x-axis enters a region where the magnetic field is along the -ve z-axis. If the charged particle continues moving along the x-axis, the magnitude of magnetic field is a. 10^3 Wb/m^2 b. 10^5Wb/m^2 c. 10^16 Wb/m^2 d. 10^-3 Wb/m^2. pls explain the solution?.
A particle of charge -16*10^-18C moving with a velocity of 10m/s along the x-axis enters a region where the magnetic field is along the -ve z-axis. If the charged particle continues moving along the x-axis, the magnitude of magnetic field is a. 10^3 Wb/m^2 b. 10^5Wb/m^2 c. 10^16 Wb/m^2 d. 10^-3 Wb/m^2. pls explain the solution? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about A particle of charge -16*10^-18C moving with a velocity of 10m/s along the x-axis enters a region where the magnetic field is along the -ve z-axis. If the charged particle continues moving along the x-axis, the magnitude of magnetic field is a. 10^3 Wb/m^2 b. 10^5Wb/m^2 c. 10^16 Wb/m^2 d. 10^-3 Wb/m^2. pls explain the solution? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A particle of charge -16*10^-18C moving with a velocity of 10m/s along the x-axis enters a region where the magnetic field is along the -ve z-axis. If the charged particle continues moving along the x-axis, the magnitude of magnetic field is a. 10^3 Wb/m^2 b. 10^5Wb/m^2 c. 10^16 Wb/m^2 d. 10^-3 Wb/m^2. pls explain the solution?.
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