Question

The self inductance of the toroidal solenoid described in the passage is

• A. $\frac{{\mu }_0{N}^{2}h}{2\pi }$
• B. $\frac{{\mu }_0{N}^{2}h}{2\pi }lo{g}_e\frac{b}{a}$
• C. $\frac{{\mu }_{0}Nh}{2\pi }lo{g}_e\frac{b}{a}$
• D. $\frac{{\mu }_0{N}^{2}h}{2\pi }lo{g}_e\frac{a}{b}$

Answer 1

The correct option is B $\frac{{\mu }_0{N}^{2}h}{2\pi }lo{g}_e\frac{b}{a}$

Consider a small section of height $h$ and width $dr$ of the total toroidal cross-section.
Using Ampere's circuital law, $\int B\cdot dl=N{\mu }_{0}i$
Or $B\times 2\pi r=N{\mu }_{0}i$
$⟹B=\frac{{\mu }_{0}iN}{2\pi r}$
Magnetic flux through this small section, $d\varphi =BdA$
where $dA=hdr$
So, ${\varphi }_{perturn}$ $=\int d\varphi =B\int dA$
${\varphi }_{perturn}$ $=\frac{{\mu }_{0}iNh}{2\pi }{\int }_a^b\frac{dr}{r}=\frac{{\mu }_{0}iNh}{2\pi }\ln \frac{b}{a}$

Total flux ${\varphi }_{total}=\text{Number of turns}\times {\varphi }_{perturn}$
${\varphi }_{total}$ $=\frac{{\mu }_0{N}^{2}hI}{2\pi }\ln \frac{b}{a}$
Using $\varphi =Li$
We get $L=\frac{{\mu }_0{N}^{2}h}{2\pi }\ln \frac{b}{a}$