Question

A toroidal solenoid with air core has an average radius of 15 cm, area of cross-section 12 cm2 and has 1200 turns. Calculate the self-inductance of the toroid. Assume the field to be uniform across the cross-section of the toroid.

Answer 1

The self inductance of a toroidal solenoid is given by L=μ0N2A2πr Here, N = number of turns A = area of cross-section r = average radius ∴L=4π×10-7×12002×12×10-42π×15×10-2 =230.4×10-5 H

Answer 2

Answer:

The self-inductance of the toroid is $\bold{230.4 \times 10^{-5} \mathrm{H}}$

Explanation:

When a wire is carrying current then the induction of a certain voltage that happens due to changing current in that wires is a form of induction called is self-induction in the given question

Number of terms is given as 1200 the radius of the solenoid is 15 cm and the area of cross section of the solenoid is given as $12 \mathrm{cm}^{2}$ to find self-induction the following formula should be used

$\bold{\mathrm{L}=\mu_{0} 2 \mathrm{A} 2 \pi \mathrm{r} \mathrm{N}}$

Applying the values we get

$\mathrm{L}=4 \pi \times 10^{-7} \times 1200 \times 12 \times 10^{-4} 2 \pi \times 15 \times 10^{-2}=230.4 \times 10^{-5} \mathrm{H}$