Question

A particle is moving in a uniform magnetic field. The charge of the particle is -5.60 nC and the magnetic field is B = -(1.25 T)Ř. A physicist measures the magnetic force on the particle to be + = -(3.40 x 10-7N)î + (7.40 x 10-7N)ſ.

1) Given the physicist’s measurement, calculate all the components of the velocity of the particle that you can.

2) Are there components of the velocity that cannot be found by the measurement of the force? Which one(s) and why?

3) Calculate the scalar (dot) product between the velocity and force vectors. What is the angle between those two vectors?

Answer 1

15 -- 3.40X10 +7.400107 B² – 1.25T Å 4 = – 5.60X10-9c know TO Thela 9 force = a (vxB) Jx B let v = verty + V₂ À v XB (veit + Vy + V2 ) x (-1.25) = 1.25 9 V j = 1.25-vy = -1.25 Vy i + 1.25V so, -1.25 Vyt + 1.15 un _ +3.40 i +7.405) x 18 7 L 10-9 - 5.60 2 C -1.25 Vy i +1.25 v = (0.607h - 1.321 ) )X10 0.607 2 وک Vy x 10 = -0.486 X10 ² -48.56 m/s 1.25 Vr 1.321 X10 2 -105.68 m/s 1.25 Vz 1) all compon onent of velocity are - 105.68 m/s =-48.56 m/s mis and we No, need time time to integrute find velocity As F = m du مره F/m v=J hat limet of integradion required.

<J TL & = -3.40X10-2 = – 105.68 2 - 48.56 5 + 7.40 x10 7 F. J = 4359.312 7 359.344)8107 - so and 92 are perpendicular to each oten Angle between then is go