To determine the velocity [tex]\( v \)[/tex] of the proton, we use the formula for the magnetic force on a charged particle moving perpendicular to a magnetic field:
[tex]\[ F = q \cdot v \cdot B \][/tex]
where
[tex]\( F \)[/tex] is the magnetic force,
[tex]\( q \)[/tex] is the charge of the proton,
[tex]\( v \)[/tex] is the velocity of the proton,
[tex]\( B \)[/tex] is the magnetic field.
We are given:
- The magnetic force, [tex]\( F = 1.8 \times 10^{-14} \)[/tex] newtons
- The charge of the proton, [tex]\( q = 1.6 \times 10^{-19} \)[/tex] coulombs
- The magnetic field, [tex]\( B = 0.025 \)[/tex] teslas
We need to solve for [tex]\( v \)[/tex]. Rearrange the formula to solve for [tex]\( v \)[/tex]:
[tex]\[ v = \frac{F}{q \cdot B} \][/tex]
Now plug in the given values:
[tex]\[ v = \frac{1.8 \times 10^{-14}}{1.6 \times 10^{-19} \cdot 0.025} \][/tex]
Evaluate the denominator first:
[tex]\[ 1.6 \times 10^{-19} \cdot 0.025 = 4.0 \times 10^{-21} \][/tex]
Now, divide the numerator by the denominator:
[tex]\[ v = \frac{1.8 \times 10^{-14}}{4.0 \times 10^{-21}} \][/tex]
[tex]\[ v = 4500000.0 \, \text{meters/second} \][/tex]
So, the velocity of the proton is [tex]\( 4.5 \times 10^6 \)[/tex] meters/second.
Therefore, the correct answer is:
E. [tex]\( 4.5 \times 10^6 \)[/tex] meters/second
