Sodium ions (Na^+) move at 0.845 m/s through a bloodstream in the arm of a perso

Question

Sodium ions (Na^+) move at 0.845 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 45.0 degtree with the motion of the sodium ions. The arm contains 80.0 cm^3 of blood with a concentration of 2.00 Times 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? Your response differs from the correct answer by more than 100%. A wire carries a current of 8.0 A in a direction that makes an angle of 33.0 degree with the direction of a magnetic field of strength 0.500 T. Find the magnetic force on a 4.00 m length of the wire.

Explanation / Answer

1.The data given in the question is,Velocity of ions: v=0.845 m/s
Magnetic field: B=0.260 T
Angle to the motion: A=45deg

Now to find the charge (q) we multiply the volume of the blood with the ions per cubic cm. And then we multiply it with the elementary charge e=1.6x10^-19 (because the sodium ions have one positive charge)

So q=V*N*e=80*2*1020 *1.6x10-19 =2560

The formula for the magnetic force is: F=B*q*v*sin(A)

F=0.260 *2560*0.845*sin(45)=397.699 N

2.Data given here,Magnetic flux (B) = 0.5 T
Current in the wire (I) = 8 A
Length of the wire (L) = 4 m
Angle of the wire () = 33º

Magnetic Force, F = B x I x L x Sin =0.5*8*4*sin(33)=8.714N

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