Question
A proton travels with a speed of 3.39×106 m/s
at an angle of 56.8
◦ with a magnetic field of
0.305 T pointed in the y direction.
The charge of proton is 1.60218 × 10−19 C.
What is the magnitude of the magnetic
force on the proton?
Answer in units of N
at an angle of 56.8
◦ with a magnetic field of
0.305 T pointed in the y direction.
The charge of proton is 1.60218 × 10−19 C.
What is the magnitude of the magnetic
force on the proton?
Answer in units of N
Answers
drwls
F = Q V x B
where the x denotes a vector cross product.
The magnitude of F is Q*V*B sin56.8
= (1.602*10^-19)*(3.39*10^6)*(0.305)*(0.837)
= _______ Newtons
where the x denotes a vector cross product.
The magnitude of F is Q*V*B sin56.8
= (1.602*10^-19)*(3.39*10^6)*(0.305)*(0.837)
= _______ Newtons