A proton is released such that its velocity of 3 Times 10^6 m/s is from right to

Question

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 proton due to the magnetic field if F = 2.4 Times 10^-16 N Find the magnitude and the direction the magnetic field B. What is the radius of the proton's path in the magnetic field? What electric field (vector!!) needs to be applied in this region if the proton is to travel undefeated through the magnetic field?

Explanation / Answer

(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

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