(12.2) Compute the following cross products. (a) 5 m/s east] × 0.4T [north- (b)
ID: 1792858 • Letter: #
Question
(12.2) Compute the following cross products. (a) 5 m/s east] × 0.4T [north- (b) 5 m/s least] × 0.47 [northeast- (c) 5 m/s [southwest] × 0.4T [north- (d) 5 m/s |southwest 0.47/northeast- (e) 5 m/s [up] × 0.47/northeast] (f) 5 m/s [west] x 0.4T 30° north of west] (g) 5 m/s [60° north of east] × 0.4T [30° west of north- (h) 5m/s (southeast] x 0.4T [down] (i) 5m/s southeast, and 60° above horizontal) × 0.4T [southeast- (j) 5 m/s [southeast, and 60° above horizontal] × 0.4T [down] (12.3) An alpha particle, with mass M.-6.64 × 10-27 kg and bearing charge qa-+2e- 3.2 × 10-19 , is injected with initial velocity to = 50000 m/s H into a region of uniform magnetic field, = 3T [O]. (e) (0) Determine the magnetic force which acts on the particle, and (B) compute the acceleration that it experiences. (b) The alpha will undergo uniform circular motion [if the region of constant field is big enough] 6) Compute the cyclotron radius (of its circular path). (ii) Compute its period (the time to complete one orbit) (ii) Compute its cyclotron frequency (the number of orbits per second)Explanation / Answer
a)
( 5 i ) x (0.4 j) = 2 k
= ( 2 mT/s ) up
b)
( 5 i ) x (0.4*(i + j)/sqrt(2))
= (2/sqrt(2)) k
= sqrt(2) k
= 1.414 m.T/s <----- up
c)
(5/sqrt(2)) *( -i - j) x (0.4 j)
= -sqrt(2) k
= 1.414 m.T/s <--------- down
d)
(5/sqrt(2))*( - i - j) x (0.4/sqrt(2)) (i + j)
= 1 (-k - k)
= -2 k
= (2 m.T/s) (down)
e)
(5 k) x ((0.4/sqrt(2)) *(i + j))
= 1.414( j - i)
= 1.414 North West
f)
(-5) i x (0.4( -cos(30 deg)i + sin(30 deg) j ))
= -2*(sin(30 deg) k)
= -1 k
= 1 m.T/s (down)
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