1. An alpha particle (q = +3.20×10 19 C) moves at a velocity of 2.45×10 4 m/s in
ID: 2060169 • Letter: 1
Question
1. An alpha particle (q = +3.20×1019C) moves at a velocity of 2.45×104m/s in the positive x
direction through a magnetic field of 0.0775 T that points in the negative z direction. What
are (a) the magnitude and (b) the direction of the magnetic force on the particle?
2. A proton is moving at 2.00×105 m/s and begins to circle at a radius of 5.00e3 m. What is the
strength of the perpendicular magnetic field acting on the proton?
3. A 15.0 ampere current flows through a 3.75 m section of wire, perpendicular to a 0.250 T
magnetic field. What is the magnitude of the force exerted by the field on the wire section?
Explanation / Answer
1.F = q*(v(cross product)B) = 3.2*10^-19*(2.45*10*4(i) (cross) 0.0775(-k)) = 6.08*10-16T (j) so, (magnitude) = 6.08*10^-16T direction = +ve y-direction 2. equating the centripetal force and the magnetic force, mv^2/r = qvB so, B= mv/qr = 1.67*10^-27*2*10^5/(1.6*10^-19*.005) = 0.4175 T 3. F = BIl = 0.25*15*3.75 = 14.06 N
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