26 (6%) Problem 3: A negative charge of q =-9.2 × 10-17C and rn = 5.3 x 10-26 kg

Question

26 (6%) Problem 3: A negative charge of q =-9.2 × 10-17C and rn = 5.3 x 10-26 kg enters a magnetic field B = 1.9 T with initial velocity v = 520 m/s, as shown in the figure. The magnetic field points into the screen KE oXX X'Xx Randomized Variables q =-9.2 × 10-17C 5.3 × 10-26 kg B=1.9T v = 520 m/s 13% Part (a) Which direction will the magnetic force be on the charge? Out of the screen.Into the screen Grade Summary To the left. OUpward To the right.Downward 0% 100% Submissions Attempts remaining: S (20% per attempt) detailed view Submit Hint I give up! Hints: 30% deduction per hint. Hints remaining: 1 Feedback: 30% deduction per feedback D 13% Part (b) Express the magnitude of the magnetic force, F, in terms of q, v , and B D 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? D 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

here,

q = -9.2 * 10^-17 C

m = 5.3 * 10^-26 kg

B = 1.9 T

v = 520 m/s

(a)

F = q * v X B

here q is negative

then the direction of magnetic force is downwards

(b)

magnitude of force = q * v * B

(c)

the magnetic force , F = q * v * B

F = 9.2 * 10^-17 * 520 * 1.9

F = 9.08*10^-14 N

(d)

in such a force , the particle moves in circular motion

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