Question

Question 1

The body contains many small currents caused by the motion of ions in the organs and cells. Measurements of the magnetic field around the chest due to currents in the heart give values of about 1 μGμG. Although the actual currents are rather complicated, we can gain a rough understanding of their magnitude if we model them as a long, straight wire.

part A

If the surface of the chest is 5.80 cmcm from this current, how large is the current in the heart?

Express your answer with the appropriate units.

II =

280μA

Part B

A magnetic field is used to suspend a wire of mass 4.0×10⁻² kgkg and length 0.13 mm . The wire is carrying a current of 13 AA . What minimum magnetic-field magnitude is needed to balance the pull of gravity?

Express your answer in teslas.

TT

Answer 1

Question 1.
We can consider the current in the heart as wires carrying currents
The magnetic field of a current-carrying wire is
B = 0 I / 2 r
Where I is current, r is the distance between the wire and the chest.
We need to find the current in the heart,
I = 2 r B / 0
B = 1 G 1 x 10-6 x 10-4 T = 10-10 T
I = 2 x 3.14 x 5.80 x 10-2 m x 10-10 T / 4 x 3.14 x 10-7
I = 2.85 x 10-5 A
Part B
The magnetic force on the wire is
F = B i L
Where B is the magnetic field, i is the current and L is the length of the wire.
This force is balanced by the weight of the wire.
F = m g
m g = Bi L
B = m g / i L
B = 4 x 10-2 kg x 9.8 m/s2 / 13 A x 0.13 x 10-3 m
B = 231.95 T