oo Verizon LTE8:44 PM 33% D Access WebAssign When a charged particle moves at an
ID: 1580647 • Letter: O
Question
oo Verizon LTE8:44 PM 33% D Access WebAssign When a charged particle moves at an angle of 12 with respect to a magnetic field, it experiences a magnetic force of magnitude F. At what angle (less than 90°) with respect to this field will this particle, moving at the same speed, experience a magnetic force of magnitude 1.6 Additional Materials Son212 A magnetic field has a magnitude of 1.20 10-3 T, and an electric field has a magnitude of 4.30 × 103 N/C. Both fields point in the same direction. A positive 1.8-HC charge moves at a speed of 3.20 106 m/s in a direction that is perpendicular to both fields. Determine the magnitude of the net force that acts on the charge. Additional Materials An ionized helium atom has a mass of 6.6 1027 kg and a speed of 9.1 105 m/s. It moves perpendicular to a 0.78-T magnetic field on a circular path that has a 0.024-m radius. Determine whether the charge of the ionized atom is te or +2e. O +e O +2e Additional Materials when beryllium-7 ions (m = 1.165 x 10-26 kg) pass through a massExplanation / Answer
force on a moving charged particle
F = q*v*B*sintheta
for theta = 12
F = q*v*B*sin12...........(1)
when force becomes 1.6 F
1.6F = q*v*B*sintheta...........(2)
2/1
1.6 = sintheta/sin12
sintheta = 1.6*sin12
theta = 19.4 degrees
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magnetic force Fb = q*v*B*sintheta = 1.8*10^-6*3.2*10^6*1.2*10^-3*sin90 = 0.006912 N
electric force Fe = E*q = 4.3*10^3*1.8*10^-6 = 0.00774 N
Fb and Fe are perpendicular
net force Fnet = sqrt(Fb^2+Fe^2)
Fnet = 0.104 N
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In the magnetic field
magnetic force provides the necessary centripetal force
Fc = Fb
m*v^2/r = q*v*B
r = m*v/qB
charge q = ne
0.024 = (6.6*10^-27*9.1*10^5)/(n*1.6*10^-19*0.78)
n = 2
charge of ion = + 2e
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