exit slit and continue to the detector My Notes C Ask You C 2/4 points I Previou
ID: 1619867 • Letter: E
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exit slit and continue to the detector My Notes C Ask You C 2/4 points I Previous Answers Knight CP3 24.XP 002. An antiproton (same properties as a proton except that a -e) is moving in the combined electric and magnetic fields of the figure below, in which 1150 V/m is the magnitude of the electric field (directed down) and 570 m/s is the antiproton's speed (to the right). B 2.5 T (a) What are the magnitude and direction of the antiproton's acceleration at this instant? x 2 down m/s 2.75e10 b) what would be the magnitude and direction of the acceleration if v were reversed? 2.57e11 x m/s up My Notes Ask You 2/2 points I Previous Answers KnightCP3 24.P033 What magnetic field strength and direction will levitate the 2.3 g wire in the figure below, in which d 11 cm? 0.137 out of page TExplanation / Answer
a)
for the magnetic field :
v = speed of the particle = 570 m/s
q = charge = 1.6 x 10-19 C
B = magnetic field = 2.5 T
magnetic force on the particle is given as
Fm = q v B Sin90
Fm = (1.6 x 10-19) (570) (2.5) Sin90
Fm = 2.3 x 10-16 N in down direction
Electric force is given as
Fe = qE = (1.6 x 10-19) (1150)
Fe = 1.84 x 10-16 N in Up direction
Fnet = net force = Fm - Fe = (2.3 x 10-16) - (1.84 x 10-16) = 4.6 x 10-17 N
m = mass = 1.67 x 10-27 kg
acceleration is given as
a = Fnet /m = 4.6 x 10-17 /( 1.67 x 10-27) = 2.75 x 1010
direction : down
b)
Fm = 2.3 x 10-16 N in Up direction
Fe = 1.84 x 10-16 N in Up direction
Fnet = net force = Fm + Fe = (2.3 x 10-16) + (1.84 x 10-16) = 4.14 x 10-16 N
m = mass = 1.67 x 10-27 kg
acceleration is given as
a = Fnet /m = 4.14 x 10-16 /( 1.67 x 10-27)
a = 2.48 x 1011 m/s2 in Up direction
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