Help! Magnetic Field Question (1 Viewer)

[tonyharrison]

HSC Paper 2008. Q11

How do you do it?

Thanks

 

[Aerath]

@ = 90*, not 40. This is because the field is perpendicular (ie, into the screen, whilst the path is parallel with the screen) to the path of the electron

Hence F = qvB = 2.688 x10 ^-14

However, F is also centripetal, so F = mv^2 / r

Sub numbers in, make r the subject, should get D.

 

[raniaaa]

use jj thomson's q/m ratio thingo:

q/m = v/rB

get the values for q and m from your data sheet

–1.602 × 10–19 / 9.109 × 10–31 = 8.0 x 10-6 / r 2.1 x 10-2

-3693270392 r = 8.0 x 10-6

r = -2.166101896 x 10-3

since r is a distance it can't be a negative, so take the absolute of it

r= 2.2 x 10-3

therefore answer is D

 

[Bunzhou]

Use this formula derived from JJ thompson

R = mv/qbsin(theta)

Where theta = 90 (perpendicular)

Sub in values and you can get radius



Derivation: F = mv^2/r = qvbsin(theta)
rearranging to make R the subject R ->
mv^2/qvbsin(theta) = mv/qbsin(theta)