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

The magnetic force on the arm is determined as 591.8 N.

How to calculate the magnetic force on the arm?

The magnetic force on the arm is calculated by applying the following formula as shown below.

F = qvB sinθ

where;

• q is the charge
• v is the speed of the charge
• B is the magnetic field strength
• θ is the angle made by the field and the velocity

The magnetic force due to one ion is calculated as;

F = (1.6 x 10⁻¹⁹) x (0.844) x (0.26) x (sin 50)

F = 2.69 x 10⁻²⁰ N

The total number of sodium ion is calculated as follows;

n = (2 x 10²⁰ ion / cm³) x (110 cm³)

n = 2.2 x 10²² ions

The total magnetic force due to all the sodium ions is calculated as;

F(total) = F × n

F(total) = 2.69 x 10⁻²⁰ N × 2.2 x 10²²

F(total) = 591.8 N

Answer 2

F = 591.8 N

Step-by-step explanation:

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

`F = q(\vec v * \vec B)`

so we can say

`F = qvBsin\theta`

here we know

`q = 1.6 * 10^(-19) C`

`v = 0.844 m/s`

`B = 0.260 T`

`\theta = 50 degree`

now we have

`F = (1.6 * 10^(-19))(0.844)(0.260)sin50`

`F = 2.69 * 10^(-20) N`

Now concentration of Sodium ions is given as

`C = 2.00 * 10^(20) ions/cm^3`

total volume of blood = 110 cc

now total number of ions is given as

`N = (2 * 10^(20))(110) = 2.2 * 10^(22) ions`

now total force on the hand is given as

`F = (2.69 * 10^(-20))(2.2 * 10^(22))`

`F = 591.8 N`