A toroidal solenoid with air core has an average radius ' $\mathrm{R}^{\prime}$, number of turns '…

A toroidal solenoid with air core has an average radius ' $\mathrm{R}^{\prime}$, number of turns ' $\mathrm{N}^{\prime}$ and area of cross-section 'A'. The self-inductance of the solenoid is (Neglect the field variation across the cross-section of the toroid)
  1. $\frac{\mu_{0} \mathrm{~N}^{2} \mathrm{~A}}{\mathrm{R}}$
  2. $\frac{\mu_{0} \mathrm{~N}^{2} \mathrm{~A}}{2 \pi \mathrm{R}}$
  3. $\frac{\mu_{0} \mathrm{NA}}{2 \pi \mathrm{R}}$
  4. $\frac{\mu_{0} \mathrm{NA}}{\mathrm{R}}$

Solution

The self-inductance $L$ of a toroidal solenoid is given by the formula: $L=\frac{\mu_0 N^2 A}{2 \pi R}$ Where: - $\mu_0$ : Permeability of free space - $N$ : Number of turns - $A$ : Cross-sectional area - $R$ : Average radius of the toroid

Asked in: MHT CET 2020 (13 Oct Shift 1)

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