Question

The heart produces a weak magnetic field that can be used to diagnose certain heart problems. It is a dipole field produced by a current loop in the outer layers of the heart.
a. It is estimated that the field at the center of the heart is 90 pT. What current must circulate around an 8.0-cm-diameter loop, about the size of a human heart, to produce this field?

Answer 1

Answer:

3.58 mA

Explanation:

Magnetic field, B = 90 pT = 90 x 10^-12 T

Diameter of loop, d = 8 cm

radius of loop, r = 4 cm = 0.04 m

Let the current be i

Magnetic field due to a current carrying loop is given by

$B =\frac{\mu _{0}}{4\pi }\times \frac{2\pi i}{r}$

$90\times 10^{-12} =10^{-7}\times \frac{2\times 3.14\times i}{0.04}$

i = 3.58 x 10^-3 A

i = 3.58 mA