Question

what is the magnetic force on a proton that is moving at 3.8×10^7 m/s up through a magnetic field that is 0.72 t and pointing toward you? the charge on a proton is 1.6×10^(-19) c. use f = qv×b sinθ.

a. 6.1×10^(-12) n left
b. 4.4×10^(-12) n right
c. 4.4×10^(-12) n left
d. 6.1×10^(-12) n right

Explanation:

Step1: Identify given values

$q = 1.6\times10^{-19}\text{ C}$, $v = 3.8\times 10^{7}\text{ m/s}$, $B=0.72\text{ T}$, $\theta = 90^{\circ}$ (since the motion of proton and magnetic - field directions are perpendicular, $\sin\theta=\sin90^{\circ} = 1$)

Step2: Apply the magnetic - force formula

$F = qvB\sin\theta$. Substitute the values:
$F=(1.6\times 10^{-19}\text{ C})\times(3.8\times 10^{7}\text{ m/s})\times(0.72\text{ T})\times1$
$F=(1.6\times3.8\times0.72)\times10^{-19 + 7}\text{ N}$
$F = 4.4\times10^{-12}\text{ N}$

Step3: Determine the direction

Using the right - hand rule: Point your fingers in the direction of the velocity (up), curl them towards the magnetic - field direction (towards you), and your thumb points to the right.

Answer:

B. $4.4\times 10^{-12}\text{ N}$ right