Question

A toroid is along coil of wire, wound over a circular core. The coefficient of self-inductance of the toroid is given by (radius $=r$ ), when the magnetic field is within it is uniform and $R>>r$

• A. $L=\frac{{\mu }_{0}N{r}^2}{2R}$
• B. $L=\frac{{\mu }_{0}Nr}{2R}$
• C. $L=\frac{{\mu }_{0}N{r}^2}{R}$
• D. $L=\frac{{\mu }_0{N}^2{r}^2}{2R}$

Answer 1

Answer: (D)

The coefficient of self-inductance of the toroid
$L=\frac{\varphi }{i},$
$\varphi =NAB$
and $B={\mu }_{0}ni$
where $n=\frac{N}{2\pi R}$
$\therefore \varphi =N.\pi r^2\left({\mu }_0\frac{N}{2\pi R}i\right)$
$\varphi =\frac{{\mu }_0{N}^2{r}^{2}i}{2R}$
$L=\frac{\varphi }{i}=\frac{{\mu }_0{N}^2{r}^2}{2R}$