Abood
- 14
- 3
- Homework Statement
- A proton travels with a speed of 3.0*10^6 m/s at an angle of 37 (degrees) west of north. A magnetic field of 0.30 T points to the north. Determine the following:
a) the magnitude of the magnetic force on the proton.
b) The direction of the magnetic force on the proton.
c) The proton's acceleration as it moves through the magnetic field
- Relevant Equations
- (Magnetic Force equations and Newton's Second Law)
F = Bvqsinθ
F = BILsinθ
F= ma
Alternative Right Hand Rule
Given:
q = 1.6*10^-19 C
B = 0.3 T north
v = 3*10^6 m/s north-west
θ = 37 (degrees)
Solution Attempt:
a) F = Bqvsinθ = (0.3)(1.6*10^-19)(3*10^6)(sin(37)) = 8.7*10^-14 N
b) Via right hand rule, F is into the page
c) a = F/m = 8.7*10^-14/1.6 × 10^−27 = 5.4*10^13 m/s^2
When I checked my answers with the given solutions, I found that answer b is incorrect which I didn't understand why.
I pointed the proton's direction to the west, the field to the north, and found my palm going inwards (into the page) but the solution shows that it should be out of the page. Did I do something incorrectly?
q = 1.6*10^-19 C
B = 0.3 T north
v = 3*10^6 m/s north-west
θ = 37 (degrees)
Solution Attempt:
a) F = Bqvsinθ = (0.3)(1.6*10^-19)(3*10^6)(sin(37)) = 8.7*10^-14 N
b) Via right hand rule, F is into the page
c) a = F/m = 8.7*10^-14/1.6 × 10^−27 = 5.4*10^13 m/s^2
When I checked my answers with the given solutions, I found that answer b is incorrect which I didn't understand why.
I pointed the proton's direction to the west, the field to the north, and found my palm going inwards (into the page) but the solution shows that it should be out of the page. Did I do something incorrectly?