Need b-h (6%) Problem 2: A negative charge of q =-2.1 × 10-17C and m = 1.2 × 10-

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Need b-h

(6%) Problem 2: A negative charge of q =-2.1 × 10-17C and m = 1.2 × 10-26 kg enters a magnetic field B = 1.8 T with initial velocity v = 860 m/s, as shown in the figure. The magnetic field points into the screen Randomized Variables q =-2.1 x 10-17C m 1.2 x 10-26 kg B=1.8 T v = 860 m/s ©theexpertta.com 13% Part (a) Which direction will the magnetic force be on the charge? 13% Part (b) Express the magnitude of the magnetic force, F, in terms of q, v, and B - Grade Summary Deductions Potential 0% 100% Submissions Attempts remaining: 8 90 per attempt) detailed view 12 3 END DELCLEAR Submit I give up! Hints: 1 for a 0% deduction. Hints remainin : 0 Feedback: 0% deduction per feedback. The required equation is the basic equation of the magnetic force 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? 13% Part (e) Express the centripetal acceleration of the particle in terms of the force F and the mass m 13% Part (f) Calculate the magnitude of a, in meters per square second 13% Part (g) Express the radius, R, of the circular motion in terms of the centripetal acceleration a and the speed v. 13% Part (h) Calculate the numerical value of the radius R, in meters

Explanation / Answer

v = 860 m/s i

B = - 1.8 k

F = q (v x B)

F = (-2.1 x 10^-17) [ (860i) x (-1.8k)]

F = - 3.25 x 10^-14 j N

(A)direction = downward

(B) F = q( v x B) = q v B

(C) F = 3.25 x 10^-14 N

(D) force is perpendicular to the velocity,

Circular motion.

(E) a = F / m = q v b / m

(f) a = 2.71 x 10^12 m/s^2

(g) q v B = m v^2 / r

r = m v / q B

(h) r = 2.73 x 10^-7 m

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