A particle with a positive charge q= 8.6x10^-6C and a mass m= 0.45g moves with v

Question

A particle with a positive charge q= 8.6x10^-6C and a mass m= 0.45g moves with velocity v = 8m/s parallel to a wall. The distance from the particle to the wall is h=45cm. A magnetic field B is applied perpendicular to the plane of the figure. The field bends the trajectory of the particle, and eventually the particle hits the wall at a 90 degree angle. Find the sign and magnitude of the magnetic field. The field is considered positive if it points out of the plane of the figure and negative if it points into the plane.

Explanation / Answer

apply centripetal force = magnetic force

i.e mv^2/r = qvB,

where m = mass o the charged particle

v = velocity,   r = radius ,

q = charge = 1.6*10^-19 C

B = magnetic field

so now magnetic field B = mv/qr

B = 0.45*10^-3 * 8/(8.6*10^-6 * 0.45)

B = 930.23 T

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