Collision in a Magnetic Field
In a bubble-chamber photograph description, a proton moving at 4.0 x 10^6 m/s enters a region of uniform magnetic field of 0.50 T directed into the page and travels along a circular arc of radius 8.4 cm. It then strikes a stationary particle, and the two particles merge into a single track that curves with a radius of 5.6 cm in the same direction. The proton mass is 1.67 x 10^-27 kg and its charge is 1.60 x 10^-19 C. A student argues that because the merged track has a smaller radius, the merged particle must have a smaller momentum than the incoming proton.
1. Evaluate the student's argument about momentum using the relationship between radius and momentum.
Read the worked answer
For a charged particle moving perpendicular to a uniform field, the magnetic force provides the centripetal force: qvB = mv^2/r, so r = mv/(qB) = p/(qB). Radius is proportional to momentum divided by charge.
Incoming proton: p = qBr = (1.60e-19)(0.50)(0.084) = 6.7e-21 kg·m/s. Check against mv = (1.67e-27)(4.0e6) = 6.7e-21 kg·m/s, which agrees.
Merged track: if the charge is still +1e, then p = (1.60e-19)(0.50)(0.056) = 4.5e-21 kg·m/s, which is indeed smaller. But momentum must be conserved in the collision, and the target was stationary, so the merged momentum should equal the incoming momentum. The smaller radius therefore signals that the merged particle's charge is larger, not that momentum vanished. If p is conserved at 6.7e-21, then q = p/(Br) = 6.7e-21/((0.50)(0.056)) ≈ 2.4e-19 C, roughly +2e. The student's reasoning ignores the charge dependence.
2. Determine the mass of the stationary target particle, stating any assumption you make.
Read the worked answer
Assume momentum is conserved and the merged charge is +2e (consistent with the radius analysis). The merged momentum is the incoming momentum, 6.7e-21 kg·m/s, and the merged speed follows from p = Mv' where M is the total mass.
From r = Mv'/(qB) with q = 2e: v' = qBr/M. Combining with p = Mv' gives p = qBr directly, so the speed is not needed to find M. Instead use the fact that the merged particle's radius is set by p and q, which we already used.
To get mass, use the merged speed: v' = p/M, and also the kinetic energy is not conserved in a perfectly inelastic collision. The clean route is to note the merged charge is +2e, so the target was a singly charged positive particle (likely another proton or a deuteron). If the target is a proton, M = 2mp and v' = p/(2mp) = 2.0e6 m/s. The radius predicted is r = Mv'/(qB) = (3.34e-27)(2.0e6)/((3.2e-19)(0.50)) ≈ 0.042 m, which is smaller than the observed 5.6 cm, so the target is heavier than a proton. Solving r = Mv'/(2eB) with v' = p/M gives M = p/(2eBr) = 6.7e-21/((3.2e-19)(0.50)(0.056)) ≈ 7.5e-28 kg, which is less than a proton and therefore unphysical. A careful answer notes the data are inconsistent with a simple two-body merge and recommends rechecking the stated radii.