12.06 What affects the curvature of charged particles in a magnetic field?
Summary
TLDRThis transcript explains how to calculate the force on a charged particle moving in a magnetic field using the equation F = BQV, where B is the magnetic flux density, Q is the charge, and V is the velocity. It discusses the circular motion of a proton in a magnetic field, deriving the radius of curvature using both force equations and momentum principles. Additionally, it compares the paths of alpha and beta particles in the same magnetic field, illustrating how mass and charge influence their curvature. The analysis emphasizes the significance of these properties in determining particle motion.
Takeaways
- 📐 The force on a charged particle in a magnetic field is calculated using the equation F = Bqv.
- 🔄 The variables in the equation are B (magnetic flux density), q (charge), and v (velocity of the particle).
- 🌀 The equation can be derived from F = BIL by substituting I with q/t (charge per unit time) to represent velocity.
- 🔺 The angle between the velocity of the particle and the magnetic field must be considered using the sine of the angle if not perpendicular.
- ⚛️ For a proton moving at 5 x 10^6 m/s in a 0.35 T magnetic field, the radius of curvature is determined to be 0.15 m.
- ⚖️ The magnetic force acting on a charged particle can be set equal to the centripetal force (mv²/r) for circular motion.
- 🔍 An alternative formula for the radius of curvature is r = mv/(Bq), derived from equating magnetic force to centripetal force.
- 🧪 When comparing alpha and beta particles in the same magnetic field, their paths differ due to their masses and charges.
- 🧲 The radius of curvature for an alpha particle is significantly larger than that of a beta particle due to its greater mass and charge.
- 📊 The calculations show that the radius of the alpha particle's curvature is 3680 times greater than that of the beta particle.
Q & A
What is the equation used to calculate the force on a charged particle moving in a magnetic field?
-The equation is F = BQV, where F is the force, B is the magnetic flux density, Q is the charge, and V is the velocity of the particle.
How does the angle of motion affect the calculation of force on a charged particle?
-If the charged particle is not moving perpendicular to the magnetic field, the angle (theta) must be included in the equation, using the sine function.
What parameters were given for the proton in the example, and what was the goal?
-The proton was moving at a speed of 5 x 10^6 m/s in a magnetic field with a magnetic flux density of 0.35 tesla. The goal was to find the radius of curvature of its motion.
How is the magnetic force on the proton calculated?
-The magnetic force is calculated using F = BQV, where B is 0.35 tesla, Q is 1.6 x 10^-19 C (the charge of a proton), and V is the velocity.
What relationship exists between magnetic force and centripetal force in this context?
-The magnetic force acting on the proton equals the centripetal force required for circular motion, which can be expressed as F = mv^2/r.
What was the calculated radius of curvature for the proton, and how was it derived?
-The radius of curvature was calculated to be 0.15 meters by equating the magnetic force to the centripetal force and rearranging the equation.
What is the derived equation for radius in a magnetic field and why is it useful?
-The derived equation is r = mv/(BQ). It is useful because it directly relates the radius of curvature to the momentum of the particle and the properties of the magnetic field.
How do the trajectories of alpha particles and beta particles differ in the same magnetic field?
-Alpha particles, which are heavier and have a greater charge, have a larger radius of curvature than beta particles, which are lighter and have a smaller charge.
What is the relative mass of an electron compared to a proton?
-An electron's mass is approximately 1/1840th of a proton's mass.
How is the ratio of the radius of curvature between alpha and beta particles calculated?
-The ratio is calculated by considering the mass and charge of both particles. The ratio results in the alpha particle having a radius of curvature that is 3680 times larger than that of the beta particle.
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