Two protons move with uniform circular motion in the presence of uniform magnetic fields.Proton one moves twice as fast as proton two.The magnitude of the magnetic field in which proton one is immersed is twice the magnitude of the magnetic field in which proton two is immersed.The radius of the circle around which proton one moves is r.What is the radius of the circle around which proton two moves?

Question

Quizplus · Accepted Answer

The radius of the circular path for a charged particle in a magnetic field is given by $r = \frac{mv}{qB}$, where $m$ is the mass of the particle, $v$ is its velocity, $q$ is its charge, and $B$ is the magnetic field strength. For proton one, moving twice as fast in a field twice as strong, both the velocity and the magnetic field strength in the formula increase by a factor of 2, which cancels out, leaving the radius unchanged. Therefore, the radius for proton two remains $r$.

Two Protons Move with Uniform Circular Motion in the Presence