After being accelerated to a speed of 1.59?10^5 m/s , the particle enters a unif

Question

After being accelerated to a speed of 1.59?10^5 m/s , the particle enters a uniform magnetic field of strength 1.00T and travels in a circle of radius 33.0 cm (determined by observing where it hits the screen as shown in the figure). The results of this experiment allow one to find m/q. Find the ratio m/q for this particle. Express your answer numerically in kilograms per coulomb.

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Explanation / Answer

Let m = mass of particle, q = charge, v = speed, B = magentic field strength, r = radisu of curvature. The particle is subject to force due to the magnetic field: F = qvB Which is balanced by a centripital force Fc = mv^2/r So equating the two: F = Fc ---> qvB = mv^2/r ---> re-arrange to get m/q m/q = rB/v = 0.33 m* 1T/1.59x10^5 m/s = 2.07x10^-6 kg/C

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