Answer to Question #95668 in Electricity and Magnetism for Julius

Question

The magnetic field of the brain has been measured to be approximately 3pT. Even though complicated, the currents that cause the field can be obtained by modelling them as a single circular loop 16 cm in diameter. What current is needed to produce such a field at the center of the loop?

Answer 1

For a thin wires we can find the magnetic induction at the center of circular loop using the formula

$$\left| {\vec B} \right| = \frac{{{\mu _0}I}}{{2R}}$$

thus we can express the current

$$I = 2\frac{{R\left| {\vec B} \right|}}{{{\mu _0}}}$$

where ${\mu _0} \approx 1.257 \cdot {10^{ - 6}}[\frac{{\text{H}}}{{\text{m}}}]$ is magnetic constant. Now let's do the calculations

$$I \approx 2 \cdot \frac{{16[{\text{cm}}] \cdot 3[{\text{pT}}]}}{{1.257 \cdot {{10}^{ - 6}}[\frac{{\text{H}}}{{\text{m}}}]}} = 2 \cdot \frac{{0.16[{\text{m}}] \cdot 3 \cdot {{10}^{ - 12}}[{\text{T}}]}}{{1.257 \cdot {{10}^{ - 6}}[\frac{{\text{H}}}{{\text{m}}}]}} \approx 7.64 \cdot {10^{ - 7}}[{\text{A}}]$$