James3shin
Diamond Member
I remember a thread involving the physics of a bike that was pulled by a wire and which direction the bike went...So i have another physics problem that Im trying to make sure I did properly for Homework.
"A proton travels with a speed of 3.00x10^6 m/s North. If the mag. field is .300T at an angle of 37 degrees N of W, solve for:
a) the magnitude and direction of the force on the proton.
b) the magnitude and direction of he acceleration of the proton."
For part "a," I used the equation F=q*V*B*Sin(theta). Where q=1.6x10^-19 C, with theta being 53 degrees, the other variables (V and B) were already given. The answer I obtained was 3.83x10^-20 N out of the page(via "right-hand rule").
For part "b," I set F=q*V*B*Sin(theta) equal to F=m*a. I then solved for acceleration and set the mass of the proton to 1.67x10^-27 Kg. The answer I obtained was 2.30x10^14 m/s^2.
Did I do these two questions properly? Thank to anyone that can help! 😀
"A proton travels with a speed of 3.00x10^6 m/s North. If the mag. field is .300T at an angle of 37 degrees N of W, solve for:
a) the magnitude and direction of the force on the proton.
b) the magnitude and direction of he acceleration of the proton."
For part "a," I used the equation F=q*V*B*Sin(theta). Where q=1.6x10^-19 C, with theta being 53 degrees, the other variables (V and B) were already given. The answer I obtained was 3.83x10^-20 N out of the page(via "right-hand rule").
For part "b," I set F=q*V*B*Sin(theta) equal to F=m*a. I then solved for acceleration and set the mass of the proton to 1.67x10^-27 Kg. The answer I obtained was 2.30x10^14 m/s^2.
Did I do these two questions properly? Thank to anyone that can help! 😀
