Question
If the outer coil of the previous problem is rotated through 90° about a diameter, what would be the magnitude of the magnetic field B at the centre?
✓
Answer
Outer Circle$\text{n}=100,\ \text{r}=100\text{m}=0.1\text{m}$
$\text{i}=2\text{A}$
$\overrightarrow{\text{B}}=\frac{\text{n}\mu_0\text{i}}{2\text{a}}=\frac{100\times4\pi\times10^{-7}\times2}{2\times0.1}=4\pi\times10^{-4}$ horizontally towards West.
Inner Circle
$\text{r}=5\text{cm}=0.05\text{m},\ \text{n}=50\text{m},\ \text{i}=2\text{A}$
$\overrightarrow{\text{B}}=\frac{\text{n}\mu_0\text{i}}{2\text{r}}=\frac{4\pi\times10^{-7}\times2\times50}{2\times0.05}=4\pi\times10^{-4}$ downwards
Net $\text{B}=\sqrt{(4\pi\times10^{-4})^2+(4\pi\times10^{-4})^2}$
$=\sqrt{32\pi^2\times10^{-8}}$
$=17.7\times10^{-4}\approx18\times10^{-4}$
$=1.8\times10^{-3}=1.8\text{mT}$
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