Question

We have seen that the heart produces a magnetic field, and that this can be used to diagnose problems with the heart. The magnetic field of the heart is a dipole field that is produced by a loop current in the outer layers of the heart. Suppose the field at the center of the heart is 90 pT (a pT is 10−12T ) and that the heart has a diameter of approximately 12 cm. What current circulates around the heart to produce this field?

Answer 1

Answer:

Current circulating is 8.59×10^-6A

Explanation:

Magnetic field at the center of the loop is given by, B= μI/2R

I = 2RB/μ

given that B is 90x 10^-12 T

radius is 0.12 m /2 = 0.06m

we know that μ is 4π x 10^-7 T.A/m

Substituting the given values we get,

I= (((2(0.06m)(90x 10^-12 T))/(4π x 10^-7 T.A/m))

I is 8.59 x 10^-6 A

Answer 2

The current circulating around the heart is 8.59×10⁻⁶A

Calculating the current:

The magnetic field B at the center of the loop of radius R with current I is given by,

B= μ₀I/2R

Rearranging the terms we get:

I = 2RB/μ₀

Given that B is 90×10⁻¹² T

radius is R = 0.12 m /2 = 0.06m

Substituting the given values we get,

I = 2×0.06×90×10⁻¹² / 4π×10⁻⁷

I = 8.59 x 10⁻⁶ A

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