The figure shows motion of a charge particle in a constant magnetic field. In wh

Question

The figure shows motion of a charge particle in a constant magnetic field. In which diagram the particle experiences the smallest magnitude of the magnetic force? (a) (b) (c) Both (a) and (c) None of these A proton moving westward with a velocity of 5.0 time sign 10^3 m/s enters a magnetic field of 0.20 T pointing southward. What is the magnitude and direction of the force that acts on the proton? 1.1 time sign 10^-16 N eastward 4.4 time sign 10^-16 N westwards 0N 1.6 time sign 10^-16 N downwards 1.6 time sign 10^-16 N upwards An electron, moving north, enters a magnetic field of certain strength. Because of this field the electron curves upward. What is the direction of the magnetic field? Downward upward towards the east towards the north towards the west A charged particle is moving horizontally eastward with a velocity of 3 time sign 10^6 m/s in a region where there is a magnetic field of magnitude 5 time sign 10^-5 T directed westward. The particle experiences a force of 15 time sign 10^-16 N northward. What is the charge on the particle? +1.0 time sign 10^-15 C -1.0 time sign 10^-15 C +1.0 time sign 10^-17 C -1.0 time sign 10^-17 C this situation cannot happen. A proton (e = 1.6 time sign 10^-19 C) is traveling east with an instantaneous velocity of 2.5 time sign 10^5 m/s when it enters a uniform magnetic field B of 0.45 T that points north. What are the magnitude and direction of the force on the electron? 1.8 time sign 10^-14 N up 1.5 time sign 10^-14 N down 1.5 time sign 10^-14 N up 1.0 time sign 10^-14 N down 1.0 time sign 10^-14 N up

Explanation / Answer

A) F=qvBsin(theta)

in figure b..... the theta=0 deg===>F=0

B)F=qvB sin(theta)

theta=90deg,

F=(1.6*10^-19)*(5*10^3)*0.2=1.6*10^-16

from right hand rule direction is upwards.

C)towads west

D)F=qvB

(15*10^-16)=q*(3*10^6)*(5*10^-5)=1*10^-17C

E)F=qvB===>F=(1.6*10^-19)*(2.5*105)*(0.45)=1.8*10^-14 up

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