A circular loop of radius a is made of single form of thin conducting ...
To find the self-inductance of the circular loop when the number of turns is increased from 1 to 8, we can use the formula for the self-inductance of a solenoid:
L = (μ₀ * n² * A) / l
Where:
L is the self-inductance
μ₀ is the permeability of free space
n is the number of turns
A is the cross-sectional area
l is the length of the solenoid
In this case, we have a circular loop, so the length of the solenoid (l) is equal to the circumference of the loop, 2πa.
When there is only 1 turn, the self-inductance is L₁.
L₁ = (μ₀ * n₁² * A₁) / l₁
When there are 8 turns, the self-inductance is L₂.
L₂ = (μ₀ * n₂² * A₂) / l₂
We want to find the ratio of L₂ to L₁:
(L₂ / L₁) = [(μ₀ * n₂² * A₂) / l₂] / [(μ₀ * n₁² * A₁) / l₁]
Canceling out μ₀ and simplifying:
(L₂ / L₁) = (n₂² * A₂ * l₁) / (n₁² * A₁ * l₂)
Since the cross-sectional area of the loop remains the same, A₂ = A₁.
(L₂ / L₁) = (n₂² * l₁) / (n₁² * l₂)
Plugging in the values, n₁ = 1 and n₂ = 8:
(L₂ / L₁) = (8² * l₁) / (1² * l₂)
(L₂ / L₁) = (64 * l₁) / l₂
So, the self-inductance when the number of turns is increased from 1 to 8 is 64L. Therefore, the answer is (a) 64L.