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## 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

[tex] F_{B}=qvBsin\theta[/tex]

## The Attempt at a Solution

First I found charge by doing some mathematical operations:

[tex] 100 cm^{3} * \frac{3*10^{20} Na+ ions}{cm^{3}}[/tex]

[tex]= 3*10^22 Na+ ions * (1.60*10^{-19})[/tex]

[tex]=4800 C [/tex] <--- I'm not sure if this procedure is right.

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