Answer
(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 $37.7^{\circ}$
Work Step by Step
(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 $37.7^{\circ}$
