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 proton moving westward with a velocity of 5.0 times 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? 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? A charged particle is moving horizontally eastward with a velocity of 3 times 10^6 m/s in a region where there is a magnetic field of magnitude 5 times10^-5 T directed westward. The particle experiences a force of 15 times 10^16 N northward. What is the charge on the particle? A proton (e = 1.6 times 10^-19 C) is traveling east with an instantaneous velocity of 2.5 times 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?

Explanation / Answer

1) F =qvB sin(theta))

sin(30) = 0.5 , sin(90) =1, sin(150) = 0.5

Correct option is (D)

2) v = -5*10^3 i m/s, B = -0.2j T , q =1.6*10^-19C

F =qvxB

F = (1.6*10^-19)((-5*10^3 i)x(-0.2j))

F = 1.6*10^-16 k N

F =1.6*10^-16 N upwards

Correct option is (E)

3) k = -(jxi)

B is along east ward

Correct option is (c)

4) F = q(vxB)

- (5*10^-5)i= q ((3*10^6 i)x(-5*10^-5 i))

this situaton cannot happend

Correct option is (E)

5) F =q(vxB)

F = (1.6*10^-19)((2.5*10^5 i)x(0.45j))

F = 1.8*10^-14 N up

Correct option is (A)

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