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A beam of protons enters a uniform magnetic field of 0.3 Tesla with a velocity of 4 x 10⁵ m/sec at an angle of 60° to the field. The radius of the helical path taken by the beam is
- a)47.1 mm
- b)32 mm
- c)18 mm
- d)46 mm
Correct answer is option 'A'. Can you explain this answer?
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Verified Answer
A beam of protons enters a uniform magnetic field of 0.3 Tesla with a ...
v1 is responsible for horizontal motion of proton
v2 is responsible for circular motion of proton
Mass of proton = 1.67 * 10-27 Kg
Charge on proton= 1.6 * 10-19 C
∴ mv22 / r = qv2 B
r = mv2 /qB
= 1.67 * 10-27 * 4 x 10⁵ / 1.6 * 10-19 * 0.3
= 0.013 m
Pitch of helix = v1 x T
Where T = 2πr / v2
= 2πr / v sin θ
⇒ Pitch of helix = v cos θ x 2πr / v sin θ
= 2 πr cotθ
= 2x 3.14 x 0.013 x cot 60°
= 0.0471 m
= 47.1 mm
Most Upvoted Answer
A beam of protons enters a uniform magnetic field of 0.3 Tesla with a ...
In order to determine the path of the protons in the magnetic field, we need to consider the Lorentz force experienced by a charged particle moving in a magnetic field.
The Lorentz force (F) experienced by a charged particle moving in a magnetic field is given by the equation:
F = qvBsinθ
where q is the charge of the particle, v is the velocity of the particle, B is the magnetic field strength, and θ is the angle between the velocity vector and the magnetic field vector.
In this case, the charged particle is a proton, which has a charge of +1.6 x 10^-19 C. The velocity of the proton is 4 x 10^6 m/s, and the magnetic field strength is 0.3 Tesla. The angle between the velocity vector and the magnetic field vector is 60 degrees.
Plugging in these values into the equation, we have:
F = (1.6 x 10^-19 C) * (4 x 10^6 m/s) * (0.3 Tesla) * sin(60 degrees)
Simplifying the equation:
F = (1.6 x 10^-19 C) * (4 x 10^6 m/s) * (0.3 Tesla) * √3/2
F = (1.6 x 4 x 0.3 x 10^-19 x 10^6 x √3/2) C*m/s*T
F = (1.92 x 10^-13 x √3) N
The direction of the Lorentz force is perpendicular to both the velocity vector and the magnetic field vector. In this case, the force will be directed towards the center of the circle.
Therefore, the proton will move in a circular path with a radius determined by the magnitude of the Lorentz force and the mass of the proton. The formula for the radius of the circular path is:
r = mv / (qB)
where m is the mass of the proton, v is the velocity of the proton, q is the charge of the proton, and B is the magnetic field strength.
The mass of a proton is approximately 1.67 x 10^-27 kg. Plugging in the values:
r = (1.67 x 10^-27 kg) * (4 x 10^6 m/s) / (1.6 x 10^-19 C * 0.3 Tesla)
Simplifying the equation:
r = (1.67 x 4 x 10^-27 x 10^6) / (1.6 x 0.3) m
r = (6.68 x 10^-21) / (0.48) m
r = 1.39 x 10^-20 m
Therefore, the proton will move in a circular path with a radius of approximately 1.39 x 10^-20 meters.
The Lorentz force (F) experienced by a charged particle moving in a magnetic field is given by the equation:
F = qvBsinθ
where q is the charge of the particle, v is the velocity of the particle, B is the magnetic field strength, and θ is the angle between the velocity vector and the magnetic field vector.
In this case, the charged particle is a proton, which has a charge of +1.6 x 10^-19 C. The velocity of the proton is 4 x 10^6 m/s, and the magnetic field strength is 0.3 Tesla. The angle between the velocity vector and the magnetic field vector is 60 degrees.
Plugging in these values into the equation, we have:
F = (1.6 x 10^-19 C) * (4 x 10^6 m/s) * (0.3 Tesla) * sin(60 degrees)
Simplifying the equation:
F = (1.6 x 10^-19 C) * (4 x 10^6 m/s) * (0.3 Tesla) * √3/2
F = (1.6 x 4 x 0.3 x 10^-19 x 10^6 x √3/2) C*m/s*T
F = (1.92 x 10^-13 x √3) N
The direction of the Lorentz force is perpendicular to both the velocity vector and the magnetic field vector. In this case, the force will be directed towards the center of the circle.
Therefore, the proton will move in a circular path with a radius determined by the magnitude of the Lorentz force and the mass of the proton. The formula for the radius of the circular path is:
r = mv / (qB)
where m is the mass of the proton, v is the velocity of the proton, q is the charge of the proton, and B is the magnetic field strength.
The mass of a proton is approximately 1.67 x 10^-27 kg. Plugging in the values:
r = (1.67 x 10^-27 kg) * (4 x 10^6 m/s) / (1.6 x 10^-19 C * 0.3 Tesla)
Simplifying the equation:
r = (1.67 x 4 x 10^-27 x 10^6) / (1.6 x 0.3) m
r = (6.68 x 10^-21) / (0.48) m
r = 1.39 x 10^-20 m
Therefore, the proton will move in a circular path with a radius of approximately 1.39 x 10^-20 meters.
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A beam of protons enters a uniform magnetic field of 0.3 Tesla with a velocity of 4 x 10 m/sec at an angle of 60° to the field. The radius of the helical path taken by the beam isa)47.1 mmb)32 mmc)18 mmd)46 mmCorrect answer is option 'A'. Can you explain this answer?
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A beam of protons enters a uniform magnetic field of 0.3 Tesla with a velocity of 4 x 10 m/sec at an angle of 60° to the field. The radius of the helical path taken by the beam isa)47.1 mmb)32 mmc)18 mmd)46 mmCorrect answer is option 'A'. Can you explain this answer? for IIT JAM 2024 is part of IIT JAM preparation. The Question and answers have been prepared according to the IIT JAM exam syllabus. Information about A beam of protons enters a uniform magnetic field of 0.3 Tesla with a velocity of 4 x 10 m/sec at an angle of 60° to the field. The radius of the helical path taken by the beam isa)47.1 mmb)32 mmc)18 mmd)46 mmCorrect answer is option 'A'. Can you explain this answer? covers all topics & solutions for IIT JAM 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A beam of protons enters a uniform magnetic field of 0.3 Tesla with a velocity of 4 x 10 m/sec at an angle of 60° to the field. The radius of the helical path taken by the beam isa)47.1 mmb)32 mmc)18 mmd)46 mmCorrect answer is option 'A'. Can you explain this answer?.
A beam of protons enters a uniform magnetic field of 0.3 Tesla with a velocity of 4 x 10 m/sec at an angle of 60° to the field. The radius of the helical path taken by the beam isa)47.1 mmb)32 mmc)18 mmd)46 mmCorrect answer is option 'A'. Can you explain this answer? for IIT JAM 2024 is part of IIT JAM preparation. The Question and answers have been prepared according to the IIT JAM exam syllabus. Information about A beam of protons enters a uniform magnetic field of 0.3 Tesla with a velocity of 4 x 10 m/sec at an angle of 60° to the field. The radius of the helical path taken by the beam isa)47.1 mmb)32 mmc)18 mmd)46 mmCorrect answer is option 'A'. Can you explain this answer? covers all topics & solutions for IIT JAM 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A beam of protons enters a uniform magnetic field of 0.3 Tesla with a velocity of 4 x 10 m/sec at an angle of 60° to the field. The radius of the helical path taken by the beam isa)47.1 mmb)32 mmc)18 mmd)46 mmCorrect answer is option 'A'. Can you explain this answer?.
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