Search any question & find its solution
Question:
Answered & Verified by Expert
A toroid is a long coil of wire wound over a circular core. If 'r' and 'R' are the radii
of the coil and toroid respectively, the coefficient of self-induction of the toroid is
(The magnetic field in it is uniform and $\mathrm{R}>>\mathrm{r}$ )
$\left(\mathrm{N}=\right.$ number of turns of the coil and $\mu_{0}=$ permeability of free space)
Options:
of the coil and toroid respectively, the coefficient of self-induction of the toroid is
(The magnetic field in it is uniform and $\mathrm{R}>>\mathrm{r}$ )
$\left(\mathrm{N}=\right.$ number of turns of the coil and $\mu_{0}=$ permeability of free space)
Solution:
2607 Upvotes
Verified Answer
The correct answer is:
$\frac{\mu_{0} \mathrm{~N}^{2} \mathrm{R}^{2}}{2 \mathrm{R}}$
$\mathrm{L}=\frac{\phi}{\mathrm{I}}, \phi=\mathrm{NAB}$
$\mathrm{B}=\mu_{0} \mathrm{nI}$
where $n=\frac{N}{2 \pi R}$
$\therefore \phi=\mathrm{N} \pi \mathrm{r}^{2}\left(\mu_{0} \frac{\mathrm{N}}{2 \pi \mathrm{R}} \mathrm{I}\right)$
$\phi=\frac{\mu_{0} N^{2} r^{2} I}{2 R}$
$\mathrm{L}=\frac{\phi}{\mathrm{I}}=\frac{\mu_{0} \mathrm{~N}^{2} \mathrm{r}^{2}}{2 \mathrm{R}}$
$\mathrm{B}=\mu_{0} \mathrm{nI}$
where $n=\frac{N}{2 \pi R}$
$\therefore \phi=\mathrm{N} \pi \mathrm{r}^{2}\left(\mu_{0} \frac{\mathrm{N}}{2 \pi \mathrm{R}} \mathrm{I}\right)$
$\phi=\frac{\mu_{0} N^{2} r^{2} I}{2 R}$
$\mathrm{L}=\frac{\phi}{\mathrm{I}}=\frac{\mu_{0} \mathrm{~N}^{2} \mathrm{r}^{2}}{2 \mathrm{R}}$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.