Question

Weak magnetic fields can be measured at the surface of the brain. Although the currents causing these fields are quite complicated, we can estimate their size by modeling them as a current loop around the equator of a $16cm$-diameter (the width of a typical head) sphere. What current is needed to produce a $30pT$field—the strength measured for one subject at the pole of this sphere?

Short answer

The current is needed to produce a $30pT$is$1.1\mu A$.

Step-by-step solution

Given information.

We need to find the current produced a $30pT$.

:Simplify

Each segment magnetic field applies to the loop, Therefore, that total magnetic exerted by one loop at any time $z$

$B_{loop}=\frac{{\mu }_o}{2}\frac{I{R}^2}{{\left(Z^2+R^2\right)}^{3/2}}\left(1\right)$

Here $z$is any point outside the radius of the sphere. Our target is to find the current I, We arrange equation ($2$)

$I=\frac{2B_{loop}{\left(z^2+R^2\right)}^{3/2}}{{\mu }_o{R}^2}\left(2\right)$

At the pole of the sphere, the radius is $R=z$. We use the condition into equation ($2$)

localid="1650966612853" $I=\frac{2B_{loop}{\left(R^2+R^2\right)}^{3/2}}{{\mu }_o{R}^2}=\frac{2^{5/2}{B}_{loop}R}{{\mu }_o}\left(3\right)$

The diameter of the sphere is $16cm$, so its radius is $R=8cm.$Now, plug the values for $B$and $R$to get as:

$$\begin{gathered} I=\frac{2^{5/2}{B}_{loop}R}{{\mu }_o} \\ =\frac{2^{5/2}\left(3\times {10}^{-12}T\right)\left(0.08m\right)}{4\pi \times {10}^{-7}T·m/A} \\ =1.1\times {10}^{-6}A \\ =\boxed{1.1\mu A} \end{gathered}$$