A uniform magnetic field points north; its magnitude is 3.5 T. A proton with kinetic energy 6.0 × 10-13 J is moving vertically downward in this field. What is the magnetic force acting on it?
This is what I've done. The answer is incorrect. Please help me find out why.
3.5T to the north
KE=1/2MV^2
2(6.0X10^-13J)/1.673X10^-27 = V^2
V^2 = 7.17x10^14
sqrt (7.17e14) = 2.6e7
Thennn
(1.60e-19C)(2.6e7)(3.5T)(sin90degrees)
= 1.45e-11 to the east.
To the east is correct, but my numerical answer is not. Help?
To find the magnetic force acting on the proton, you correctly calculated the speed of the proton as 2.6 x 10^7 m/s. However, there is a mistake in your calculation for the magnetic force.
The magnetic force on a moving charged particle can be calculated using the equation: F = q * v * B * sin(theta), where q is the charge of the particle, v is the velocity of the particle, B is the magnitude of the magnetic field, and theta is the angle between v and B.
In this case, the charge of the proton is q = 1.60 x 10^-19 C, the velocity of the proton is v = 2.6 x 10^7 m/s, and the magnitude of the magnetic field is B = 3.5 T. Since the particle is moving vertically downwards and the magnetic field points north, the angle theta between the velocity and the magnetic field is 90 degrees.
Substituting the values into the equation, we get:
F = (1.60 x 10^-19 C) * (2.6 x 10^7 m/s) * (3.5 T) * sin(90 degrees)
sin(90 degrees) = 1, so we can simplify the equation to:
F = (1.60 x 10^-19 C) * (2.6 x 10^7 m/s) * (3.5 T)
Multiplying these values, we get:
F = 1.456 x 10^-11 N
So, the magnetic force acting on the proton is 1.456 x 10^-11 N, directed to the east.
Please double-check your calculations to ensure there are no errors in the numerical values used.