A particle of mass 75.0 g moves in a horizontal circle (ignore gravity in this p
ID: 1335176 • Letter: A
Question
A particle of mass 75.0 g moves in a horizontal circle (ignore gravity in this problem) and has a charge of 60.0 C. The particle is connected to a string that helps confine it to the circular orbit. The particle moves in a counter-clockwise direction as viewed from above and there is a magnetic field of 2.00 T that points down. The particle radius of the particle’s orbit is 7.00 mm and it completes 1,000 orbits per second. Find the tension in the string. (Hint: be careful with conversions between grams and kilograms).
Explanation / Answer
speed = v = distance / time = 1000*2*pi*r/t = (1000*2*pi*7*10^-3)/1 = 43.98 m/s
given magnetic field = B = 2 T
the magnetic field exerts a force Fb = q*v*B*sintheta
theta = angle between the field and velocity vector = 90
Fb = q*v*B = 60*10^-6*43.98*2 = 0.0052776 N
the Fb provides the necessary centripetal force
let tension in the string is makes an angle theta with the vertical
T*sintheta = Fb
T*costheta = m*g = 75*10^-6**10^-3*9.8 = 7.35*10^-7 N
T = sqtt(Fb^2+(mg)^2)
T = sqrt(0.0052776^2+(7.35*10^-7)^2)
T = 0.00528 N <<<-----------answer
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