Question

Sodium ions (Na+) move at 0.844 m/s through a bloodstream in the arm of a person standing near a large magnet. The magnetic field has a strength of 0.260 T and makes an angle of 50.0° with the motion of the sodium ions. The arm contains 110 cm3 of blood with a concentration of 2.00 1020 Na+ ions per cubic centimeter. If no other ions were present in the arm, what would be the magnetic force on the arm?

Answer 1

Answer:

F = 591.8 N

Explanation:

Force on a moving charge in constant magnetic field is given as

$F = q(\vec v \times \vec B)$

so we can say

$F = qvBsin\theta$

here we know

$q = 1.6 \times 10^{-19} C$

$v = 0.844 m/s$

$B = 0.260 T$

$\theta = 50 degree$

now we have

$F = (1.6 \times 10^{-19})(0.844)(0.260)sin50$

$F = 2.69 \times 10^{-20} N$

Now concentration of Sodium ions is given as

$C = 2.00 \times 10^{20} ions/cm^3$

total volume of blood = 110 cc

now total number of ions is given as

$N = (2 \times 10^{20})(110) = 2.2 \times 10^{22} ions$

now total force on the hand is given as

$F = (2.69 \times 10^{-20})(2.2 \times 10^{22})$

$F = 591.8 N$