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A proton enters a magnetic field with a velocity of 1.5 x 105 m/s making an angle of 30° with the field. What is the force experienced by the proton if B = 3.0 T.
- a)3.6×10-14 N
- b)1.2x-14 N
- c)4.8x-14 N
- d)2.4x-14 N
Correct answer is option 'A'. Can you explain this answer?
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A proton enters a magnetic field with a velocity of 1.5 x 105m/s makin...
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A proton enters a magnetic field with a velocity of 1.5 x 105m/s makin...
Degrees with the direction of the field. If the magnetic field has a strength of 0.5 Tesla, what is the force experienced by the proton?
We can use the formula for the magnetic force on a charged particle:
F = qvBsinθ
where F is the force, q is the charge of the particle (in this case, the charge of a proton is +1.6 x 10^-19 C), v is the velocity of the particle, B is the magnetic field strength, and θ is the angle between the velocity and the magnetic field.
Plugging in the given values, we get:
F = (1.6 x 10^-19 C)(1.5 x 10^5 m/s)(0.5 T)sin(30°)
F = 6.0 x 10^-14 N
Therefore, the force experienced by the proton is 6.0 x 10^-14 N.
We can use the formula for the magnetic force on a charged particle:
F = qvBsinθ
where F is the force, q is the charge of the particle (in this case, the charge of a proton is +1.6 x 10^-19 C), v is the velocity of the particle, B is the magnetic field strength, and θ is the angle between the velocity and the magnetic field.
Plugging in the given values, we get:
F = (1.6 x 10^-19 C)(1.5 x 10^5 m/s)(0.5 T)sin(30°)
F = 6.0 x 10^-14 N
Therefore, the force experienced by the proton is 6.0 x 10^-14 N.
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A proton enters a magnetic field with a velocity of 1.5 x 105m/s makin...
F = q(V×B)= q(V×B sin@)
F = 1.6×10^-19C×(1.5×10^5m/s×3T sin(30•))
F = 3.6×10^-14N
F = 1.6×10^-19C×(1.5×10^5m/s×3T sin(30•))
F = 3.6×10^-14N
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A proton enters a magnetic field with a velocity of 1.5 x 105m/s making an angle of 30° with the field. What is the force experienced by the proton if B = 3.0 T.a)3.6×10-14Nb)1.2x-14Nc)4.8x-14Nd)2.4x-14NCorrect answer is option 'A'. Can you explain this answer? for Class 12 2026 is part of Class 12 preparation. The Question and answers have been prepared according to the Class 12 exam syllabus. Information about A proton enters a magnetic field with a velocity of 1.5 x 105m/s making an angle of 30° with the field. What is the force experienced by the proton if B = 3.0 T.a)3.6×10-14Nb)1.2x-14Nc)4.8x-14Nd)2.4x-14NCorrect answer is option 'A'. Can you explain this answer? covers all topics & solutions for Class 12 2026 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A proton enters a magnetic field with a velocity of 1.5 x 105m/s making an angle of 30° with the field. What is the force experienced by the proton if B = 3.0 T.a)3.6×10-14Nb)1.2x-14Nc)4.8x-14Nd)2.4x-14NCorrect answer is option 'A'. Can you explain this answer?.
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