Answer
(a) $B = 0$
(b) $B = 0.16~T$ in the $+z$ direction
(c) $B = 0.040~T$ in the $-z$ direction
Work Step by Step
We can use the Biot-Savart law to solve this question.
(a) $B = \frac{\mu_0~q~v~sin~\theta}{4~\pi~r^2}$
$B = \frac{(4\pi\times 10^7~H/m)~(1.6\times 10^{-19}~C)~(1.0\times 10^7~m/s)~sin~0^{\circ}}{(4~\pi)~(0.01~m)^2}$
$B = 0$
(b) $B = \frac{\mu_0~q~v~sin~\theta}{4~\pi~r^2}$
$B = \frac{(4\pi\times 10^7~H/m)~(1.6\times 10^{-19}~C)~(1.0\times 10^7~m/s)~sin~90^{\circ}}{(4~\pi)~(0.01~m)^2}$
$B = 0.16~T$
By the right hand rule, the magnetic field is in the $+z$ direction.
(c) $B = \frac{\mu_0~q~v~sin~\theta}{4~\pi~r^2}$
$B = \frac{(4\pi\times 10^7~H/m)~(1.6\times 10^{-19}~C)~(1.0\times 10^7~m/s)~sin~90^{\circ}}{(4~\pi)~(0.02~m)^2}$
$B = 0.040~T$
By the right hand rule, the magnetic field is in the $-z$ direction.
