Question

If the linear dimensions of the core of a cylindrical coil are doubled, the inductance of the coil will be (assuming complete winding over the core)

• A. Doubled
• B. Four fold
• C. Eight times
• D. Remains unchanged

Answer 1

Answer: (C)

Eight times

Answer 2

The inductance of a coil is given as L = $\frac{\mu_{0}N^{2}A}{\mathcal{l}}$ where N = total number of turns of the coil; A = area of cross section of the coil = π r²; r = radius of the core of the coil, = length of the coil. If is doubled the total number of turns will be doubled

⇒ $\frac{L_{2}}{L_{1}}\mspace{6mu} = \mspace{6mu}\left( \frac{N_{2}}{N_{1}} \right)^{2}\mspace{6mu}\left( \frac{A_{2}}{A_{1}} \right)\mspace{6mu}\left( \frac{\mathcal{l}_{1}}{\mathcal{l}_{2}} \right)\mspace{6mu} = \mspace{6mu}\left( \frac{N_{2}}{N_{1}} \right)^{2}\mspace{6mu}\left( \frac{\pi r_{2}^{2}}{\pi r_{1}^{2}} \right)\mspace{6mu}\left( \frac{\mathcal{l}_{1}}{\mathcal{l}_{2}} \right)$⇒

$\frac{L_{2}}{L_{1}}\mspace{6mu} = \mspace{6mu}(2)^{2}(2)^{2}\mspace{6mu}\left( \frac{1}{2} \right)\mspace{6mu} = \mspace{6mu} 8 (\because\mspace{6mu}\frac{N_{2}}{N_{1}}\mspace{6mu} = \mspace{6mu}\frac{r_{2}}{r_{1}}\mspace{6mu} = \mspace{6mu}\frac{\mathcal{l}_{2}}{\mathcal{l}_{1}}\mspace{6mu} = \mspace{6mu} 2)$.

Hence (3) is correct.