A stationary proton is in a uniform magnetic field of 0.20 T. What is the magnit
ID: 1544557 • Letter: A
Question
A stationary proton is in a uniform magnetic field of 0.20 T. What is the magnitude of the magnetic force on the proton? zero 3.2 times 10^-20 N 3.2 times 10^-21 N 1.6 times 10^-21 N 1.6 times 10^-20 N A proton is projected with a velocity of 7.0 times 10^3 m/s into a magnetic field of 0.60 T perpendicular to the motion of the proton. What is the force that acts on the proton? 0 N 13 times 10^-16 N 3.4 times 10^-16 N 6.7 times 10^-16 N 4.2 times 10^-16 N An electron traveling due north with speed 4.0 times 10^5 m/s enters a region where the Earth's magnetic field has the magnitude 5.0 times 10^-5 T and is directed downward at 45 degree below horizontal. What force acts on the electron? 3.2 times 10^-19 N 3.2 times 10^-18 N 2.3 times 10^-19 N 2.3 times 10^-18 N 2.3 times 10^-20 N A proton moving eastward with a velocity of 5.0 times 10^3 m/s enters a magnetic field of 0.20 T pointing northward. What is the magnitude and direction of the force that acts on theExplanation / Answer
1 proton at stationary v = 0 m/s , B = 0.20 T
magnetic force on proton is f = qvB sin theta = q*0*B sin theta = 0 N
answer A)
2) proton charge q = 1.6*10^-19 C, velocity v = 7*10^3 m/s
magnetic field B = 0.60 T , theta = 90 degrees
magnetic force f = qVB sin theta
f = 1.6*10^-19*7000*0.60 sin90 N = 6.72*10^-16 N
answer D)
3. electron travelling north with speed v = 4*10^5 m/s
B = 5*10^-5 T, angle theta = -45 degrees
the force on electron is
f = qVB sin theta
f = -1.6*10^-19*400000*5*10^-5 sin(-45) N = 2.26274*10^-18 N = 2.3*10^-18 N
answer D)
4.
proton charge q = 1.6*10^-19 C, velocity v = 5*10^3 m/s towards east
magnetic field B = 0.20 T north direction , theta = 90 degrees
f = 1.6*10^-19*5000*0.20 sin90 N = 1.6*10^-16 N
the direction is upwards
answer A)
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