Question

Magnetoencephalography (MEG) is a technique for measuring changes in the magnetic field of the brain caused by external stimuli such as touching the body or viewing images of food. Such a change in the field occurs due to electrical activity (current) in the brain. During the process, magnetic sensors are placed on the skin to measure the magnetic field at that location....

Magnetoencephalography (MEG) is a technique for measuring changes in the magnetic field of the brain caused by external stimuli such as touching the body or viewing images of food. Such a change in the field occurs due to electrical activity (current) in the brain. During the process, magnetic sensors are placed on the skin to measure the magnetic field at that location. Typical field strengths are a few femtoteslas (1 femtotesla = 1 = 10-15 T). An adult brain is about 140 mm wide, divided into two sections (called "hemispheres" although the brain is not truly spherical) each about 70 mm wide. We can model the current in one hemisphere as a circular loop, 65.0 mm in diameter, just inside the brain. The sensor is placed so that it is along the axis of the loop 2.30 cm from the center. A reasonable magnetic field is 5.35 at the sensor. According to this model, what is the current in this hemisphere of the brain? Number What is the magnetic field at the center of the hemisphere of the brain? Number

Answer 1

PART A)

$\:$

The magnitude of the magnetic field due to a circular loop current carrying coil of radius R is given by the equation,

$\:$

$B=\frac{\mu _{o}iR^{2}}{2(x^{2}+R^{2})^{3/2}} \: \: \: \: \: \: \: \: \: (1)$

$\:$

here x represent the distance away from the center along the axis of the loop. So, rearraging eq. (1) for the current,

$\:$

$i= \frac{2B(x^{2}+R^{2})^{3/2}}{\mu _{o}R^{2}}$

$\:$

since we want the current in one hemisphere of the brain, use

$\:$

$R=R_{brain}=\frac{70mm}{2}=35mm$

$\:$

thus, isnerting data given,

$\:$

$i= \frac{2(5.35\times 10^{-15}T)\left [ (2.30cm)^{2}+(35mm)^{2} \right ]^{3/2}}{(4\pi \times 10^{-7}N/A^{2})(35mm)^{2}}$

$\:$

or

$\:$

${\color{Blue} i=0.051060408 \times 10^{-8} \, A \approx 5.11\times 10^{-10}\, A}$

$\:$

PART B)

$\:$

In this case use,

$\:$

$R=R_{loop}=\frac{65mm}{2}=32.5mm$

$\:$

then plug in this value in eq. (1) plus the current calculated in PART A)

$\:$

$B=\frac{(4\pi \times 10^{-7}N/A^{2})(5.11\times 10^{-10}A)(32.5mm)^{2}}{2\left [ (2.30cm)^{2}+(32.5mm)^{2} \right ]^{3/2}}$

$\:$

or

$\:$

${\color{Blue} B=537.304957571 \times 10^{-17} \, T\approx 5.37\times 10^{-15}\, T=5.37 \, fT}$