Answer to Question #95668 in Electricity and Magnetism for Julius

Question #95668

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?


1
Expert's answer
2019-10-02T09:52:25-0400

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


B=μ0I2R\left| {\vec B} \right| = \frac{{{\mu _0}I}}{{2R}}


thus we can express the current


I=2RBμ0I = 2\frac{{R\left| {\vec B} \right|}}{{{\mu _0}}}

where μ01.257106[Hm]{\mu _0} \approx 1.257 \cdot {10^{ - 6}}[\frac{{\text{H}}}{{\text{m}}}] is magnetic constant. Now let's do the calculations


I216[cm]3[pT]1.257106[Hm]=20.16[m]31012[T]1.257106[Hm]7.64107[A]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}}]


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