MHT CET · Physics · Electromagnetic Induction
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)
- A \(\frac{2 \mu_{0} \mathrm{r}^{2}}{\mathrm{~N}^{2} \mathrm{R}}\)
- B \(\frac{\mu_{0} \mathrm{~N}^{2} \mathrm{R}^{2}}{2 \mathrm{r}}\)
- C \(\frac{\mu_{0} \mathrm{~N}^{2} \mathrm{R}^{2}}{2 \mathrm{R}}\)
- D \(\frac{\mu_{0} \mathrm{R}}{2 \mathrm{~N}^{2} \mathrm{r}^{2}}\)
Answer & Solution
Correct Answer
(C) \(\frac{\mu_{0} \mathrm{~N}^{2} \mathrm{R}^{2}}{2 \mathrm{R}}\)
Step-by-step Solution
Detailed explanation
\(\mathrm{L}=\frac{\phi}{\mathrm{I}}, \phi=\mathrm{NAB}\)
\(\mathrm{B}=\mu_{0} \mathrm{nI}\)
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 \(n=\frac{N}{2 \pi R}\)
\(\therefore \phi=\mathrm{N} \pi \mathrm{r}^{2}\left(\mu_{0} \frac{\mathrm{N}}{2 \pi \mathrm{R}} \mathrm{I}\right)\)
\(\phi=\frac{\mu_{0} N^{2} r^{2} I}{2 R}\)
\(\mathrm{L}=\frac{\phi}{\mathrm{I}}=\frac{\mu_{0} \mathrm{~N}^{2} \mathrm{r}^{2}}{2 \mathrm{R}}\)
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