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A beam of protons traveling at 1.20 km/s enters a uniform
Chapter 27, Problem 24E(choose chapter or problem)
A beam of protons traveling at \(1.20 \mathrm{~km} / \mathrm{s}\) enters a uniform magnetic field, traveling perpendicular to the field. The beam exits the magnetic field, leaving the field in a direction perpendicular to its original direction (Fig. E27.24). The beam travels a distance of 1.18 cm . What is the magnitude of the magnetic field?
Equation Transcription:
Text Transcription:
1.20 km/s
Questions & Answers
QUESTION:
A beam of protons traveling at \(1.20 \mathrm{~km} / \mathrm{s}\) enters a uniform magnetic field, traveling perpendicular to the field. The beam exits the magnetic field, leaving the field in a direction perpendicular to its original direction (Fig. E27.24). The beam travels a distance of 1.18 cm . What is the magnitude of the magnetic field?
Equation Transcription:
Text Transcription:
1.20 km/s
ANSWER:
Step 1 of 2
The expression for magnetic field in this question can be calculated by using the equation …..(1), where m is the mass of the proton, v is the speed of proton, q is the charge of a proton and r is the radius of the circular path.
Given,
Mass of proton
Charge of proton
Now, the beam travels 1.18 cm in the magnetic field. If it has to travel a whole circular path, total distance traveled would be