**Example.**

Newton’s second law of motion is given as the following formula.
$$F=m cdot a. tag1$$
Here we use standard notation for force $F$, for the mass of observed object $m$, and the object’s acceleration $a$.

#

This law states that the force acting on an object is directly proportional to the product of the object’s mass and acceleration. What we want to show with this law is that it is necessary to know at least two variables to determine the force acting on the observed object. In our case, we must know the mass of the observed object and its acceleration.

**Note.** The mass of an object is expressed in kilograms while the acceleration is in $mathrm{m/s^2}$.

– We conclude that Newton’s second law (observed as a formula) helps us to understand the motion of the observed object.

$$
F = Bqv
$$

$$
kgcdot m/s^2 = (T)(Acdot s)(m/s)
$$

$$
kgcdot m/s^2 = dfrac{(Acdot s)cdot T cdot m}{s}
$$

$$
kgcdot dfrac{m}{s^2 cdot m} = (A)cdot T
$$

$$
dfrac{kg }{s^2 cdot A} = T
$$

$$
dfrac{kg }{s^2 cdot A} = T
$$

$textbf{underline{textit{Knowns:}}}$

The magnetic force exerted on a charged particle due to an external magnetic field is given by the following equation
$F = q v B tag{1}$

And it is given that the charge of the proton is $1.6times 10^{-19} ~ rm{A.s}$, which is moving with a speed of $2.4 times 10^{5}$ m/s, and the magnitude of the applied magnetic field through which the proton is moving is 4.5 T.

$$
textbf{underline{textit{Solutions}}}
$$

enumerate[bfseries (a)]
item Substituting the givens in equation (1), we get the value of the force on the proton
begin{align*}
F &= left( 1.6 times 10^{-19} ~ rm{A ~ s}right) left( 2.4 times 10^{5}~ rm{m/s} right) left( 4.5 ~ rm{T}right)\
&= 1.728 times 10^{-13} ~ dfrac{rm{A}~ rm{s} ~ rm{m} ~ rm{T}}{rm{s}}\
&= 1.728 times 10^{-13} ~ {rm{A} ~ rm{m} ~ rm{T}}
intertext{And, the unit of Tesla is equivalent to $dfrac{rm{kg}}{rm{s}^2 rm{A}}$, thus we have}
&= 1.728 times 10^{-13} ~ {rm{A} ~ rm{m}} ~ dfrac{rm{kg}}{rm{s}^2 ~ rm{A}}\
intertext{Thus, the final unit of the calculated force is}
&= 1.728times 10^{-13} ~ rm{kg} ~ dfrac{rm{m}}{rm{s}^2}\
intertext{Which is equivalent unit to mass times acceleration which is the definition of the force and since the units in SI, thus this unit of force is in units of Newton, hence the units of the R.H.S of equation (1) have the same physical dimension as that of force, therefor the units are correct.}
end{align*}

enumerate[bfseries (b)]
item Substituting in equation (1) by the givens and to estimate the order of the magnitude of the givens we get
begin{align*}
F&= 1.6 times 10^{-19} times 2.4 times 10^{5} times 4.5 \
&= left( 16 times 2.4 times 4.5 right) times 10^{-14}
end{align*}
We find that the order of the magnitude of the answer when written in scientific notation $m times 10^{n}$ is $n=-14$

enumerate[bfseries (c)]
item Comparing the estimated order of magnitude from part(b), to the calculated answer from part (a) we find that the order of magnitude is 10 times less than the expected value, where $m=17.28$ .\

enumerate[bfseries (a)]
item $1.73 times 10^{-13}$ N.
item $n=-14$.
item Expected order of magnitude is 10 times less than the expected value.
item 4 significant figures.

Known: $F=Bqv$
Unkown: $v=?$

$$
begin{align*}
Bqv &= F tag{Reflexive property of equality} \
dfrac{Bq}{Bq} v &= dfrac{F}{Bq} tag{Dividing both sides with $Bq$} \
v &= dfrac{F}{Bq} tag{$Bq$ cancel out on left side of the equation}
end{align*}
$$

$$
v=dfrac{F}{Bq}
$$

No, you would not discard 9.801 $m/s^{2}$ as the accepted value for acceleration due to gravity. That value was determined by many other experiments over hundreds of years. In order to discard that value, you must prove what the other experiments have done wrong this entire time.

It is more likely that there were imprecise measurements in your experiment instead. Maybe your way to recording time was not accurate, or friction and air resistance was neglected.