A positively charged particle of mass 7.21E-8 kg is traveling dueeast with a spe

Question

A positively charged particle of mass 7.21E-8 kg is traveling dueeast
with a speed of 85.3 m/s and enters a 0.340 T uniform magneticfield.
The particle moves through one-quarter of a circle in a time of2.30E-3
s, at which time it leaves the field heading due south. All duringthe
motion the particle moves perpendicular to the magnetic field. Whatis
the magnitude of the magnetic force acting on the particle?

Explanation / Answer

the magnetic force is the centripetal force : => F = M*V^2/R The particle moves through one-quarter of a circle in a time of2.30E-3 s => the period(move 1 circle) : T = 4*2.3x10^-3 =9.2x10^-3 s as we know : T = 2R/V => R = T*V/2 => the force :   F = M*V^2/(T*V/2) =2*M*V/T = 2*7.21x10^-8*85.3/9.2x10^-3 = 4.2x10^-3 N

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