A proton is released such that its velocity of 3 times 106 m

A proton is released such that its velocity of 3 times 10^6 m s is from right to left across this page The proton enters a region of uniform magnetic field (B). such that it deflected toward the bottom of this page If the magnitude of the magnetic force on the proton due to the magnetic field is 2.4 times 10^-6 N Find the magnitude und the direction the magnetic field B. What is the radius of the proton\'s path in the magnetic field

Solution

(a)The magnetic force acting on a charged particle moving through a magnetic field B is,

F = Bqv sin(theta)

B = F / qv sin90

B = 2.4*10^-16 / ( 1.6*10^-19*3*10^6 m/s )

B = 0.5*10^-3 T

B = 0.5 mT

From right rule, the direction of magnetic field is into the page.

b) q v B = mv^2/r

r = m v/(q B)

= 1.67E-27*3.0E6/(1.6E-19*5.0E-4)

=62.63 m

c) E is up

q E = q v B

E = v B

= 3.0E6*5.0E-4

= 1500 N/C up