(25%) Problem 1: A negative charge of q= -3.7 x 10-17c and m = 1.7 x 10-26 kg en

Question

(25%) Problem 1: A negative charge of q= -3.7 x 10-17c and m = 1.7 x 10-26 kg enters a magnetic field B = 19T with initial velocity v = 780 m/s, as shown in the figure. The magnetic field points into the screen. Randomized Variables x x x x x x x x0x x x x x x x x x X X X q=-3.7 x 10-17 m = 1.7 x 10-26 kg B = 1.9T y = 780 m/s x x x x x x x X X X X X X X ©theexpertta.com 2 / 13% Part (a) Which direction will the magnetic force be on the charge? 2 / 13% Part (b) Express the magnitude of the magnetic force, F, in terms of q, v, and B. / 13% Part (c) Calculate the magnitude of the force F, in newtons. * * 13% Part (d) Under such a magnetic force, which kind of motion will the charge undergo? A 13% Part (e) Express the centripetal acceleration of the particle in terms of the force F and the mass m. DA 13% Part (f) Calculate the magnitude of a, in meters per square second. 4 13% Part (g) Express the radius, R, of the circular motion in terms of the centripetal acceleration a and the speed v. > 4 13% Part (h) Calculate the numerical value of the radius R, in meters. R = 1 Grade Summary Deductions 0% Potential 100% sin() cotan() atan() cos() asin() acotan() tanh0 tan) t ( 7 8 9 HOME acos() 1^ AJ 4 5 6 sinh() * 1 2 3 Submissions Attempts remaining: 3 (4% per attempt) detailed view cosh(0. Cotanh)

Explanation / Answer

d) Since the magnetic force is perpendicular to the velocity, the particle will undergo circular motion

e) Centripetal Acceleration = Magnetic Force / mass

a = F / m ; F=qvB

f) a = qvB / m = 3.2255*1012 m/s2

g) a = v2 / R ====> R = v2 / a

h) R = v2 / a ===> R = 1.8862*10-7 meters

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