Question

The magnetic field around the head has been measured to be approximately$\mathbf{3}\mathbf{.}\mathbf{0}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{8}}\mathit{G}$. Although the currents that cause this field are quite complicated, we can get a rough estimate of their size by modelling them as a single circular current loop 16 cm (the width of a typical head) in diameter. What is the current needed to produce such a field at the center of the loop?

Short answer

The current needed to produce$3.0\times {10}^{-8}G$ at the center of the loop is$3.8\times {10}^{-7}A$_.

Step-by-step solution

The magnetic field at the center of the loop

The magnetic field at the center of the loop is given by:

$B=\frac{{\mu }_{0}I}{2R}$

Where B is the magnetic field due to wire,${\mu }_0$ is the permeability of the vaccum, l is the current through the wire, and R is the distance from the wire.

The calculation of the current flowing in the loop 

Given data:

• The magnetic field at the center of the loop is 3.0 ×10^-12 T_.

Using formula :

$\begin{array}{r}B=\frac{{\mu }_{0}I}{2R}\\ I=\frac{2BR}{{\mu }_0}\end{array}$

Now, putting the values of constants in the above equation as:

$\begin{array}{r}I=\frac{2\times \left(8\times {10}^{-2}\right)\times \left(3.0\times {10}^{-12}\right)}{4\times \pi \times {10}^{-7}}\\ I=3.8\times {10}^{-7}\mathrm{A}\end{array}$

Thus, the current needed to produce$3.0\times {10}^{-8}\mathrm{G}$ at the center of the loop is$3.8\times {10}^{-7}\mathrm{A}$_.