A wire of resistance \( 4\,\Omega \) is used to make a coil of radius \( 7\,\mathrm{cm} \). The wire has a diameter of \( 1.4\,\mathrm{mm} \) and the resistivity of its material is \( 2 \times 10^{-7}\,\Omega\,\mathrm{m} \). The number of turns in the coil will be
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \( 70 \)
First find the length of the wire from its resistance, then divide by the length of one turn. The resistance of a wire is
\( \displaystyle R = \frac{\rho\,l}{A} \;\Rightarrow\; l = \frac{R\,A}{\rho} \)
\( r_{\mathrm{wire}} = 0.7\,\mathrm{mm} = 0.7\times10^{-3}\,\mathrm{m} \)
\( \displaystyle A = \pi r_{\mathrm{wire}}^{2} = \pi\,(0.7\times10^{-3})^{2} = 1.54\times10^{-6}\,\mathrm{m^2} \)
\( \displaystyle l = \frac{4 \times 1.54\times10^{-6}}{2\times10^{-7}} = 30.8\,\mathrm{m} \)
Each turn is a circle of radius \( 7\,\mathrm{cm} \) (the thickness of the wire is neglected), so its length is
\( \displaystyle 2\pi R_{\mathrm{coil}} = 2\pi \times 0.07 = 0.44\,\mathrm{m} \)
\( \displaystyle N = \frac{30.8}{0.44} = 70 \)