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
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
To find the magnitude of the magnetic force on the proton, you can use the formula for the magnetic force on a charged particle moving in a magnetic field:
F = q * v * B * sinθ
Where:
F is the magnetic force
q is the charge of the particle (in this case, the charge of a proton, which is 1.60218 × 10^(-19) C)
v is the velocity of the particle (3.39×10^6 m/s)
B is the magnetic field strength (0.305 T)
θ is the angle between the velocity vector and the magnetic field vector (56.8°)
Let's plug in the values and calculate the magnitude of the magnetic force:
F = (1.60218 × 10^(-19) C) * (3.39×10^6 m/s) * (0.305 T) * sin(56.8°)
You can use a calculator to determine the sine of 56.8° and then calculate the final result.