Answer
(a) $81^{\circ}$
(b) $38^{\circ}$
(c) $1.2^{\circ}$
Work Step by Step
(a) We know that
$F=Bqvsin\theta$
$\implies sin\theta=\frac{F}{Bqv}$
$\implies \theta=sin^{-1}(\frac{F}{Bqv})$
We plug in the known values to obtain:
$\theta=sin^{-1}(\frac{4.8\times 10^{-6}N}{(0.95T)(16m/s)(0.32\times 10^{-6}C)})$
$\theta=81^{\circ}$
(b) $\theta=sin^{-1}(\frac{F}{Bqv})$
We plug in the known values to obtain:
$\theta=sin^{-1}(\frac{3.0\times 10^{-6}N}{(0.95T)(16m/s)(0.32\times 10^{-6}C)})$
$\theta=38^{\circ}$
(c) $\theta=sin^{-1}(\frac{F}{Bqv})$
We plug in the known values to obtain:
$\theta=sin^{-1}(\frac{1.0\times 10^{-7}N}{(0.95T)(16m/s)(0.32\times 10^{-6}C)})$
$\theta=1.2^{\circ}$
