What is the Correct Angle for the Net Magnetic Force?

AI Thread Summary
The net magnetic force acting on a charged particle moving in a magnetic field is calculated to be 9.77 x 10^-3 N. The components of the force are Fx ~ 0.003708 N and Fy ~ 0.009034 N. The angle calculation for the net force with respect to the +x axis is misinterpreted; simply using the inverse tangent of Fy/Fx does not yield the correct angle. The correct approach involves recognizing that the force is perpendicular to both the velocity and magnetic field vectors, necessitating the use of the right-hand rule for accurate direction determination. Understanding these vector relationships is key to resolving the angle correctly.
lovelyrwwr
Messages
48
Reaction score
0
One component of a magnetic field has a magnitude of 0.0266 T and points along the +x axis, while the other component has a magnitude of 0.0648 T and points along the -y axis. A particle carrying a charge of +2.96 × 10-5 C is moving along the +z axis at a speed of 4.71 × 103 m/s. (a) Find the magnitude of the net magnetic force that acts on the particle. (b) Determine the angle that the net force makes with respect to the +x axis.


I got the correct answer for part A, which is simply 9.77x10-3 N.
Fx ~ 0.003708 N
Fy ~ 0.009034 N
F = sqrt(Fx^2 + Fy^2)
F = 9.77x10-3 N

However, I am not getting the correct answer for B.

I thought it was simply:
theta = InverseTangent(Fy/Fx) = 67.68 deg but this wasn't correct.

Then I thought I had to add 90 to get theta = 67.68 + 90 = 157.68 deg since F is perpendicular to both field (B) and velocity (V), but this was not correct either. What am I doing incorrectly? Thank you!
 
Last edited:
Physics news on Phys.org
According to the vector product (hope you know this version) the force on the positive charge
is perpendicular to both the velocity vector and the magnetic field vector:

\vec{F} = q \vec{v} \times \vec{B}

, that is you turn the velocity vector towards the magnetic field vector. The force vector points in the direction in which a righthand screw advances due to this turning. This means \vec{F} will be perpendicular to the plane containing the two vectors \vec{v} and \vec{B}. Or you may know the right-hand rule for determining the direction of \vec{F}.
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top