A) \[\frac{{{\mu }_{0}}{{N}_{0}}{{I}_{0}}}{4({{R}_{2}}-{{R}_{1}})}\ln ({{R}_{2}}/{{R}_{1}})\]
B) \[\frac{2{{\mu }_{0}}{{N}_{0}}{{I}_{0}}}{{{R}_{2}}-{{R}_{1}}}\ln ({{R}_{2}}/{{R}_{1}})\]
C) \[\frac{{{\mu }_{0}}{{N}_{0}}{{I}_{0}}}{2({{R}_{2}}-{{R}_{1}})}\ln {{({{R}_{2}}/{{R}_{1}})}^{2}}\]
D) \[\frac{{{\mu }_{0}}{{N}_{0}}{{I}_{0}}}{2({{R}_{2}}-{{R}_{1}})}\ln ({{R}_{2}}/{{R}_{1}})\]
Correct Answer: D
Solution :
Consider a small strip of thickness \[dr\] at distance \[r\] from the centre, then number of turns in this strip would be \[dN=\left( \frac{{{N}_{0}}}{{{R}_{2}}-{{R}_{1}}} \right)dr\] Magnetic field due to this element at the centre of the coil will be \[dB=\frac{{{\mu }_{0}}dN{{I}_{0}}}{2r}=\frac{{{\mu }_{0}}{{N}_{0}}{{I}_{0}}}{2({{R}_{2}}-{{R}_{1}})}\frac{dr}{r}\] Total magnetic field at the centre of the spiral \[B=\int_{r={{R}_{1}}}^{r={{R}_{2}}}{\frac{{{\mu }_{0}}{{N}_{0}}{{I}_{0}}}{2({{R}_{2}}-{{R}_{1}})}\frac{dr}{r}=\frac{{{\mu }_{0}}{{N}_{0}}{{I}_{0}}}{2({{R}_{2}}-{{R}_{1}})}\ln \left( \frac{{{R}_{2}}}{{{R}_{1}}} \right)}\]You need to login to perform this action.
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