Question

An adult brain is about 140 mm wide and divided into two sections (called "hemispheres" although the brain is not truly spherical), which are each about 70 mm wide. Model the current in one hemisphere as a circular current loop 65.0 mm in diameter. A sensor is placed so that it is along the axis of the loop, 2.10 cm from the center. Suppose a magnetic field of 4.15 fT is measured by the sensor. According to the model, what is the current I in the hemisphere of the brain being measured

Answer 1

Answer: 3.61×10^5 A

Step-by-step explanation: Since the brain has been modeled as a current carrying loop, we use the formulae for the magnetic field on a current carrying loop to get the current on the hemisphere of the brain.

The formulae is given below as

B = u×Ia²/2(x²+a²)^3/2

Where B = strength of magnetic field on the axis of a circular loop = 4.15T

u = permeability of free space = 1.256×10^-6 mkg/s²A²

I = current on loop =?

a = radius of loop.

Radius of loop is gotten as shown... Radius = diameter /2, but diameter = 65mm hence radius = 32.5mm = 32.5×10^-3 m = 3.25×10^-2m

x = distance of the sensor away from center of loop = 2.10 cm = 0.021m

By substituting the parameters into the formulae, we have that

4.15 = 1.256×10^-6 × I × (3.25×10^-2)²/2{(0.021²) + (3.25×10^-2)²}^3/2

4.15 = 13.2665 × 10^-10 × I/ 2( 0.00149725)^3/2

4.15 = 1.32665 ×10^-9 × I / 2( 0.000058)

4.15 × 2( 0.000058) = 1.32665 ×10^-9 × I

I = 4.15 × 2( 0.000058)/ 1.32665 ×10^-9

I = 4.80×10^-4 / 1.32665 ×10^-9

I = 3.61×10^5 A