# A toroidal solenoid with an air core has an average radius of 0.15 m, area of cross section 12 × 10^(-4) m^2 and 1200 turns…

Q: (a) A toroidal solenoid with an air core has an average radius of 0.15 m, area of cross section 12 × 10-4 m2 and 1200 turns. Obtain the self inductance of the toroid. Ignore field variation across the cross section of the toroid. (b) A second coil is 300 turns is wound closely on the toroid above. If the current in the primary coil is increased from zero to 2.0 A in 0.05 s, obtain the induced emf in the secondary coil.

Sol: (a) $\large B = \mu_0 n_1 I = \frac{\mu_0 N_1 I}{l}$

Total Magnetic Flux $\large \phi_B = N_1 B A = \frac{\mu_0 N_1^2 I A}{l}$

$\large \phi_B = \frac{\mu_0 N_1^2 I A}{2 \pi r}$

But φB = L I

$\large L = \frac{\mu_0 N_1^2 A}{2 \pi r}$

$\large L = \frac{4\pi \times 10^{-7}\times (1200)^2 \times 12 \times 10^{-4}}{2\pi \times 0.15}$

= = 2.3 × 10-3 H = 2.3 mH

(b) $\large |e| = \frac{d}{dt}(\phi_2)$ ; where φ2 is the total magnetic flux linked with the second coil.

$\large |e| = \frac{d}{dt}(N_2 B A)$