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{\phi}{i}, $
$\phi=N A B $
and $B=\mu_{0} n i $
where $n=\frac{N}{2 \pi R} $
$\therefore \phi=N . \pi r^{2}\left(\mu_{0} \frac{N}{2 \pi R} i\right) $
$\phi=\frac{\mu_{0} \,N^{2}\, r^{2} i}{2 R} $
$L=\frac{\phi}{i}=\frac{\mu_{0} N^{2} r^{2}}{2 R}$