kmarks Window Help webassign.net EXAMPLE 19.2 A Proton Moving in a Magnetic Fiel
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kmarks Window Help webassign.net EXAMPLE 19.2 A Proton Moving in a Magnetic Field GOAL Calculate the magnetic force and acceleration when a particle moves at an angle other than 90° to the field. PROBLEM A proton moves at 8.00 x 106 m/s along the x- axis. It enters a region in which there is a magnetic field of magnitude 2.50 T, directed at an angle of 60.0° with the x-axis and lying in the xy-plane (see figure). (a) Find the initial magnitude and direction of the magnetic force on the proton. (b) Calculate the proton's initial acceleration. 60 STRATEGY Finding the magnitude and direction of the magnetic force requires substituting values into the equation The mapneic force Fon a proton a in the posetive z for magnetic force, F = qvB sine, and using the right-hand rule, de tion whenVandBle l, the ry-plane. Applying Newton's second law solves part (b). SOLUTION (A) Find the magnitude and direction of the magnetic force on the proton. Substitute v = 8.00 x 106 m/s, the magnetic field strength B = 2.50 T, the angle, and the charge of a proton into the magnetic force equation F-919 sin -( 1.60 x 10-19 C)(8.00 x 106 m/s)(2.50 Tosin 600) F 2.77 x 10-12 N Apply right-hand rule number 1 to find the initial direction of the magnetic force Point the fingers of the right hand in the x direction (the direction of 0) and then curl them toward B. The thumb points upward, in the positive z-direction. (B) Calculate the proton's initial acceleration. Substitute the force and the mass of a proton into Newton's second law: ma = F (1.67 x 1027 kg)a " 2.77 x 10-12 N u 1.66 x 1015 m/s? acBook AirExplanation / Answer
a) f = qvB sin(theta)
f = 1.6*10^-19*8.05*10^6*2.36sin60 = 26.32*10^-14 N
b) f = ma
a = f/m = 26.32*10^-14/1.67*10^-27 = 15.76*10^13 m/s^2
excersice:
a = f/m = 26.32*10^-14/9.11*10^-31 = 2.89*10^17 m/s^2
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