A) \[\frac{{{\mu }_{0}}nI}{(b-a)}{{\log }_{e}}\frac{a}{b}\]
B) \[\frac{{{\mu }_{0}}nI}{2(b-a)}\]
C) \[\frac{2{{\mu }_{0}}nI}{b}\]
D) \[\frac{{{\mu }_{0}}nI}{2(b-a)}{{\log }_{e}}\frac{b}{a}\]
Correct Answer: D
Solution :
Consider an element of thickness\[dr\]at a distance\[r\]from the centre of spiral coil. Number of turns in coil\[=n\] Number of turns per unit length \[=\frac{n}{b-a}\] Number of turns in element\[dr=dn\] Number of turns per unit length in element\[dr\] \[=\frac{ndr}{b-a}\] \[ie\], \[dn=\frac{ndr}{b-a}\] Magnetic field at its centre due to element \[dr\]is \[dB=\frac{{{\mu }_{0}}Idn}{2r}=\frac{{{\mu }_{0}}I}{2}\frac{n}{(b-a)}\frac{dr}{r}\] \[\therefore \] \[B=\int_{a}^{b}{\frac{{{\mu }_{0}}Indr}{2(b-a)r}=\frac{{{\mu }_{0}}In}{2(b-a)}\int_{a}^{b}{{}}\frac{dr}{r}}\] \[=\frac{{{\mu }_{0}}In}{2(b-a)}{{\log }_{e}}\left( \frac{b}{a} \right)\]You need to login to perform this action.
You will be redirected in
3 sec