Question
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?
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?
Answers
Related Questions
A uniform magnetic field points North; its magnitude is 1.5 T. A proton with kinetic energy 8.0x10–1...
A 42 mT magnetic field points due west. If a proton of kinetic energy 9x10E-12 J enters this field i...
A 42 mT magnetic field points due west. If a proton of kinetic energy 9x10E-12 J enters this field i...