Question

. (a) Calculate the self-inductance of a toroidal coil consisting of 1000 turns of wire, each of radius 1 cm wound uniformly on a non-ferromagnetic ring of mean radius 10 cm. (10)

(b) A circuit containing an air-cored solenoid of self-inductance 0.1 H carries a steady current of 1 A. A switch in the circuit is opened, reducing the current to zero in a time of 1 ms. Determine the magnitude of the induced emf across the solenoid, assuming the current fallsto zero at a constant rate.

Answer 1

(a)

$L = \frac{\mu_0 N^2A}{2γ} \\ \mu_0 = 4 \pi \times 10^{-7}\;H/m \\ Area=A=\pi r^2 \\ r=radius = 10 \;cm = 0.1 \;m \\ N= number \;of \;turns = 1000 \\ L = \frac{4 \pi \times 10^{-7} \times (1000)^2 \times (0.1)^2}{2 \times 0.1} \\ L=0.062\;H$

(b)

Initial current

I₁ = 1 A

Final current

I₂=0

Time interval

t = 1ms $= 10^{-3} \;s$

Self-inductance of a solenoid

$L= \frac{\mu_0N^2A}{l}$

Magnetic flux

$Ø = BA \\ = \frac{\mu_0NI}{l}A \\ = \frac{\mu_0N^2A}{l} \times \frac{I}{N} \\ Ø = L \times \frac{I}{N} \\ E_{ind}=-N\frac{ΔØ}{Δt} \\ = -N \frac{L}{N} \times \frac{ΔI}{Δt} \\ = - \frac{L(I_2-I_1)}{Δt} \\ E_{ind} = - \frac{0.1(0-2)}{10^{-3}} = 100 \;V$