A 10 g bullet carrying a charge of 4.00 is fired horizontally at a...

The magnetic force acting on the bullet is perpendicular to its velocity and is given by $F = qvB$.

The acceleration of the bullet due to this magnetic force is $a = \dfrac{F}{m} = \dfrac{qvB}{m}$.

Since the deflection is very small, the horizontal velocity remains approximately constant. The time taken to travel a horizontal distance $x$ is $t = \dfrac{x}{v}$.

The deflection $y$ caused by the magnetic field is given by the second equation of motion:

$y = \dfrac{1}{2}at^2 = \dfrac{1}{2} \left(\dfrac{qvB}{m}\right) \left(\dfrac{x}{v}\right)^2 = \dfrac{qBx^2}{2mv}$

Substituting the given values:
$q = 4.00 \times 10^{-6} \text{ C}$
$B = 500 \times 10^{-6} \text{ T} = 5 \times 10^{-4} \text{ T}$
$x = 100 \text{ m}$
$m = 10 \text{ g} = 10^{-2} \text{ kg}$
$v = 270 \text{ m s}^{-1}$

$y = \dfrac{4 \times 10^{-6} \times 5 \times 10^{-4} \times (100)^2}{2 \times 10^{-2} \times 270}$

$y = \dfrac{20 \times 10^{-10} \times 10^4}{5.4 \times 10^{-2} \times 10^2} = \dfrac{20 \times 10^{-6}}{5.4}$

$y = \dfrac{200}{54} \times 10^{-6} \approx 3.7 \times 10^{-6} \text{ m}$