Question

you and a few friends were to charge up your hair by rubbing a balloon against your head, you could run in a circle and create a current. What is the magnitude and the direction of the magnetic field that would be created in the middle of this circle?

Answer 1

Question #83187, Physics / Electromagnetism

you and a few friends were to charge up your hair by rubbing a balloon against your head, you could run in a circle and create a current. What is the magnitude and the direction of the magnetic field that would be created in the middle of this circle?

During friction of the rubbing a balloon, a positive charge is formed on the hair. The total charge that was formed on the hair is $\mathbf{n}$ - the number of electrons passed to by rubbing a balloon.

Current that occurs when running in a circle $I = \frac{e}{T} = N\frac{ne}{T}$ , where $N$ - number of friends, $n$ - number of electrons and $e$ - charge of electron, T-period Magnetic moment $p_m = IS, S = 2\pi r^2$ , then $p_m = N\frac{ne}{T} 2\pi r^2$

the direction of the magnetic field is shown in the figure

magnitude of the magnetic field $B = \mu_0\mu \frac{I}{2r} = \mu_0\mu N\frac{ne}{2rT}$

Let's try to estimate the magnitude of the magnetic moment

Let $-\mathrm{N} = 4, \mathrm{n} = 50, \mathrm{T} = 4 \mathrm{~s}, \mathrm{r} = 2 \mathrm{~m}$

$e = 1.6 \times 10^{-19} C, \mu_0 = 4\pi \times 10^{-7} H / m$

$p_{m} = 4 \cdot \frac{50 \cdot 1.6 \times 10^{-19}}{4} 2 \cdot 3.14 \cdot 4 = 2.1 \times 10^{-16} A \cdot m^{2}$

$B = 4\pi \times 10^{-7} \cdot 4 \cdot \frac{50 \cdot 1.6 \times 10^{-19}}{2 \cdot 2 \cdot 4} = 2.51 \times 10^{-24} W b$

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