   Chapter 19, Problem 10P

Chapter
Section
Textbook Problem

Sodium ions (Na+) move at 0.851 m/s through a blood-stream 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 51.0° with the motion of the sodium ions. The arm contains 100 cm3 of blood with a concentration of 3.00 × 1020 Na+ ions per cubic centimeter. If no other ions were present in the arm, what would be the magnetic force on the arm?

To determine
The magnetic force on the arm.

Explanation

Given info: The speed of the sodium ions is 0.851ms-1 . The strength of the magnetic field is 0.254T . The angle between the magnetic field and the motion of the ion is 51.0° . The concentration of sodium ion in blood is 3.00×1020ionscm-3 . The volume of blood in the arm is 100cm3 .

Explanation:

The magnetic force on a charged particle is given by,

F=qvBsinθ       (1)

• q is the charge on the particle
• v is the velocity of the particle
• B is the magnetic field
• θ is the angle between the velocity and the magnetic field

When there are N number of charged particles available, the magnetic force will be,

F=NqvBsinθ       (2)

The total number of ions can be found out by,

N=CV       (3)

• C is the concentration of the ions
• V is the volume

Combining (2) and (3),

F=CVqvBsinθ

Substitute 1.60×1019C for q , 0

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