Question

Two circular coils made of similar wires but of radii 20 cm and 40 cm are connected in parallel. Find the ratio of the magnetic fields at their centers.

Moderate

Solution

When the coils are connected in parallel, the ratio of currents flowing in these coils will be equal to reciprocal of the ratio of their resistances

$\therefore \frac{{i}_{1}}{{i}_{2}}=\frac{{R}_{2}}{{R}_{1}}=\frac{\rho \left({l}_{2}/A\right)}{\rho \left({l}_{1}/A\right)}\phantom{\rule{0ex}{0ex}}(asboththecoilsaremadeofsimilarwires)\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{{\ell}_{2}}{{\ell}_{1}}=\frac{2\pi {r}_{2}}{2\pi {r}_{1}}=\frac{{r}_{2}}{{r}_{1}}\therefore \frac{{B}_{1}}{{B}_{2}}=\frac{\left({\mu}_{0}{i}_{1}\right)/2{r}_{1}}{\left({\mu}_{0}{i}_{2}\right)/2{r}_{2}}\phantom{\rule{0ex}{0ex}}\therefore \frac{{B}_{1}}{{B}_{2}}=\frac{{i}_{1}}{{i}_{2}}\times \frac{{r}_{2}}{{r}_{1}}\left[\because \frac{{i}_{1}}{{i}_{2}}=\frac{{r}_{2}}{{r}_{1}}\right]\phantom{\rule{0ex}{0ex}}\therefore \frac{{B}_{1}}{{B}_{2}}=\frac{{r}_{2}^{2}}{{r}_{1}^{2}}={\left(\frac{40}{20}\right)}^{2}=\frac{4}{1}=4$

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