A particle with a mass of 4.82 times 10^27 kg traveling with a velocity v = 2.73

Question

A particle with a mass of 4.82 times 10^27 kg traveling with a velocity v = 2.73 times 10^6 m/s is placed in a magnetic field directed into the screen as shown in the diagram below. The magnetic field has a strength of B = 1.86 T and the particle has a charge of + 3.83 times 10^-16 C. What is the size of the force felt by the particle in this magnetic field? ______N What is the direction of this force when the particle is traveling in the direction shown in the diagram? Select one: Up the screen Down the screen Into the screen Out of the screen To the left To the right It has no direction as It is zero What is the acceleration of the particle? _____ms^-2 What is the radius of the circular path followed by the particle in this field? _______m

Explanation / Answer

a) according to right hand rume tipp of the fingers represents the velocity and palm gives the direction of field and thumb gives the direction of force.this is for electron

so force is acting on the proton is down the screen.

b) f = qvB = ma

a = 3.83*10^-18*2.73*10^6*1.86/4.82*10^-27

a = 4.035*10^15 m/s^2

a = v^2/r

r = 2.73^2*10^12/4.035*10^15 = 1.85*10^-3 m = 1.85 mm

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