## College Physics (4th Edition)

(a) The component of the velocity perpendicular to the magnetic field is $3.1\times 10^7~m/s$ (b) The component of the velocity parallel to the magnetic field is $4.0\times 10^7~m/s$ (c) The angle between the velocity vector and the magnetic field is $37.7^{\circ}$
(a) We can find the perpendicular component of velocity $v_{per}$: $F = q~v_{per} B$ $v_{per} = \frac{F}{q~B}$ $v_{per} = \frac{2.3 \times 10^{-12}~N}{(1.6\times 10^{-19}~C)(0.47~T)}$ $v_{per} = 3.06\times 10^7~m/s$ The component of the velocity perpendicular to the magnetic field is $3.1\times 10^7~m/s$ (b) We can find the component of the velocity parallel $v_{par}$ to the magnetic field: $v_{per}^2+v_{par}^2 = v^2$ $v_{par}^2 = v^2 - v_{per}^2$ $v_{par} = \sqrt{v^2 - v_{per}^2}$ $v_{par} = \sqrt{(5.0\times 10^7~m/s)^2 - (3.06\times 10^7~m/s)^2}$ $v_{par} = 4.0\times 10^7~m/s$ The component of the velocity parallel to the magnetic field is $4.0\times 10^7~m/s$ (c) We can find the angle between the velocity vector and the magnetic field: $sin~\theta = \frac{3.06\times 10^7~m/s}{5.0\times 10^7~m/s}$ $\theta = arcsin~(\frac{3.06\times 10^7~m/s}{5.0\times 10^7~m/s})$ $\theta = 37.7^{\circ}$ The angle between the velocity vector and the magnetic field is $37.7^{\circ}$