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Two long solenoids Sand S2 have equal lengths and the solenoid S1 is placed co-axially inside the solenoid S2. If the current in both the solenoids is doubled, then the mutual inductance of both the solenoids will become:
  • a)
    Four times
  • b)
    Double
  • c)
    Remain unchanged
  • d)
    None of these
Correct answer is option 'C'. Can you explain this answer?
Most Upvoted Answer
Two long solenoidsS1andS2have equal lengths and the solenoidS1is place...
We know that if there are two solenoids of equal length and one solenoid is placed coaxially inside the other solenoid then the mutual inductance of solenoid 1 with respect to solenoid 2 will be equal to the mutual inductance of solenoid 2 with respect to solenoid 1.
The mutual inductance of both the solenoids is given as,

Where, n1 = number of turns per unit length of solenoid 1, n2 = number of turns per unit length of solenoid 2, r1 = radius of the inner solenoid, and l = length of both the solenoids
By equation (1) it is clear that the mutual inductance of both the solenoids does not depend on the current in the solenoids.
Therefore, when the current in both the solenoids is doubled, the mutual inductance of both the solenoids S1 and S2 will remain unchanged. 
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Two long solenoidsS1andS2have equal lengths and the solenoidS1is place...
Explanation:

When two solenoids are placed coaxially, the mutual inductance between them depends on the number of turns of wire, the area of cross-section, and the relative positions of the solenoids.

The formula for mutual inductance (M) is:

M = (μ₀N₁N₂A)/(L₂)

Where,
M = Mutual inductance
μ₀ = Permeability of free space
N₁, N₂ = Number of turns in solenoid 1 and solenoid 2 respectively
A = Area of cross-section of solenoid 1
L₂ = Length of solenoid 2

In this case, both solenoids have equal lengths and solenoid S1 is placed coaxially inside solenoid S2. Let's consider the initial situation where the current in both solenoids is I.

Case 1: Initial Situation
Number of turns in S1 = N₁
Number of turns in S2 = N₂

Now, the mutual inductance (M) between S1 and S2 is:
M = (μ₀N₁N₂A)/(L₂)

Case 2: Current in both solenoids is doubled
Number of turns in S1 = N₁
Number of turns in S2 = N₂

The current in both solenoids is doubled. Therefore, the new current in each solenoid is 2I.

The mutual inductance (M') between S1 and S2 in this case is:
M' = (μ₀N₁'N₂'A)/(L₂)

where N₁' and N₂' are the new number of turns in S1 and S2 respectively.

Since the solenoid S1 is placed coaxially inside solenoid S2, the number of turns in S1 remains the same (N₁' = N₁).

Now, comparing the expressions for M and M', we can see that all the other parameters (μ₀, N₂, A, L₂) remain the same in both cases.

Therefore, M' = (μ₀N₁N₂A)/(L₂) = M

Hence, the mutual inductance of both solenoids remains unchanged when the current in both solenoids is doubled.
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Two long solenoidsS1andS2have equal lengths and the solenoidS1is placed co-axially inside the solenoidS2. If the current in both the solenoids is doubled, then the mutual inductance of both the solenoids will become:a)Four timesb)Doublec)Remain unchangedd)None of theseCorrect answer is option 'C'. Can you explain this answer?
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