Question

The self-inductance L of a toroid having circular cross-section of radius \(r\) and major radius \(R\) is (\(r\ll R\))

• A. \(\frac{\mu _{0}N^{2}R^{2}}{r^{2}}\)
• B. \(\frac{\mu _{0}N^{2}R^{2}}{2r}\)
• C. \(\frac{\mu _{0}N^{2}r^{2}}{R^{2}}\)
• D. \(\frac{\mu _{0}N^{2}r^{2}}{2R}\)

Hint:

Use the field inside a toroid and the flux linkage over its cross-section.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept
Inside a toroid with \(N\) turns and \(r\ll R\), \(B=\dfrac{\mu_{0}NI}{2\pi R}\), nearly uniform across the cross-section.

Step 2: Flux linkage
Cross-section area \(A=\pi r^{2}\). Flux linkage \(N\phi=N\cdot B\pi r^{2}=\dfrac{\mu_{0}N^{2}Ir^{2}}{2R}\).

Step 3: Inductance
\(L=\dfrac{N\phi}{I}=\dfrac{\mu_{0}N^{2}r^{2}}{2R}\).

Final Answer:
The self-inductance is mu zero N squared r squared over 2R. \[ \boxed{\text{(D) }\dfrac{\mu_{0}N^{2}r^{2}}{2R}} \]