Magnetic fields and sodium ions

[AznBoi]

Homework Statement

Sodium ions (Na+) move at 0.851 m/s through a bloodstream in the arm of a person standing near a large magnet. The magnetic field has a strength of 0.254 T and makes an angle of 51 degrees with the motion of sodium ions. The arm contains 100 cm^3 of blood with 3 x 10^20 Na+ ions per cubic centimeter. If no other ions were present in the arm, what would be the magnetic force on the arm?

Homework Equations

$$F_{B}=qvBsin\theta$$

The Attempt at a Solution

First I found charge by doing some mathematical operations:
$$100 cm^{3} * \frac{3*10^{20} Na+ ions}{cm^{3}}$$
$$= 3*10^22 Na+ ions * (1.60*10^{-19})$$
$$=4800 C$$ <--- I'm not sure if this procedure is right.

If the charge is correct, then: $$F_{b}=4800 (0.851 m/s)(0.254 T)sin 51$$ $$=806.32 N$$

 

[AznBoi]

Also, although this is irrelevant and not associated with the problem above, what is the charge of a "singly charged ion" with mass= 2.18 x 10^-26kg? Thanks.

 

[Saketh]

Also, although this is irrelevant and not associated with the problem above, what is the charge of a "singly charged ion" with mass= 2.18 x 10^-26kg? Thanks.

The assumption is that the ion has a charge of the fundamental charge, $e$ (charge on a proton or electron). This has a value of 1.60 x 10^-19 C. $Na^+$ is a singly charged ion.