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A circular loop of radius a' is made of single form of thin conducting wire. The self inductance of this loop is L'. If the no. of turns in the loop is increased from 1 to 8. the self inductance would be:-
  • a)
    64 L
  • b)
    8L
  • c)
    2√2L
  • d)
    L/8
Correct answer is option 'A'. Can you explain this answer?
Verified Answer
A circular loop of radius a is made of single form of thin conducting ...

Now if n' is changed from 1 to 8 

L" = 64 L
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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.
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A circular loop of radius a is made of single form of thin conducting wire. The self inductance of this loop is L. If the no. of turns in the loop is increased from 1 to 8. the self inductance would be:-a)64 Lb)8Lc)2√2Ld)L/8Correct answer is option 'A'. Can you explain this answer?
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