The magnetic field in a cyclotron is $0.50 \mathrm{~T}$. Find the magnitude of the magnetic force on a proton with velocity of $1.0 \times 10^{7} \mathrm{~m} / \mathrm{s}$ in a plane perpendicular to the field.
A) $8.0 \times 10^{-13} \mathrm{~N}$
B) $6.0 \times 10^{-13} \mathrm{~N}$
C) $6.0 \times 10^{-14} \mathrm{~N}$
D) $8.0 \times 10^{-14} \mathrm{~N}$

Question

Quizplus · Accepted Answer

The magnetic force on a charged particle moving in a magnetic field is given by $F = qvB\sin\theta$, where $q$ is the charge of the particle, $v$ is its velocity, $B$ is the magnetic field strength, and $\theta$ is the angle between the velocity and the magnetic field. For a proton ($q = 1.6 \times 10^{-19} \, \text{C}$), moving perpendicular to the field ($\sin\theta = 1$), with $v = 1.0 \times 10^{7} \, \text{m/s}$ and $B = 0.50 \, \text{T}$, the force is $F = (1.6 \times 10^{-19} \, \text{C})(1.0 \times 10^{7} \, \text{m/s})(0.50 \, \text{T}) = 8.0 \times 10^{-13} \, \text{N}$.

The Magnetic Field in a Cyclotron Is 0.50 T0.50 \mathrm{~T}0.50 T Find the Magnitude of the Magnetic Force on a Find

The Magnetic Field in a Cyclotron Is $0.50 \mathrm{~T}$ Find the Magnitude of the Magnetic Force on a Find