Question

need before 12:00 noon central time july 30, 2020

Problem 3 A particle with charge of 2 (uc), mass of 2X10-20 (kg) and velocity of 200 (m/s) enters into a magnetic field of 0.06 (T). The particle experiences a force of 1.01X10% (N) in the magnetic field. A) Calculate the angle between velocity and magnetic field directions. B) Calculate the acceleration of charged particle in the magnetic field. C) Draw a diagram showing direction of velocity, magnetic field, force and the angle between velocity and magnetic field, using RHR-1 D) In order to maximize the force, what should be the angle between velocity vector and magnetic field direction? What would be the value of the maximum magnetic force at this angle?

Answer 1

q = 2*10^(-6) C

m= 2*10^(-20) Kg

v= 200m/s

B =0.06 T

F = 1.01*10^(-5) N

F = qV X B = qVB sin C , V = velocity , C = angle between velocity of path and B vector

1.01*10^(-5) N = 2*10^(-6) C *200m/s*0.06T*sin(C)

C = arcsin(1.01*10^(-5)/(2*10^(-6) C *200m/s*0.06T)) = 24.9 deg

b)

F = ma

a = F/m = 1.01*10^(-5) N / 2*10^(-20) Kg = 5.05*10^(-14) m/s^2

c)

나 2 24.4 ㅂ 가 ㅋ

d)

C = 909 deg , gives sin90 = 1, max value

F(max) = 2*10^(-6) C *200m/s*0.06T = 2.4*10^(-5) N

4

A)

V = 0.5 A * 4 Ohms = 2 V

B)

Current direction: clockwise in loop

C)

B = $\mu _{0}I/2\pi r$ = 4*pi*10^(-7) H/m *0.5A / ( 2*pi*0.20m) = 5*10^(-7) T

D)

Direction of B = Into the computer screen