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For a solenoid keeping the turn density ...

For a solenoid keeping the turn density constant its length makes halved and its cross section radius is doubled then the inductance of the solenoid increased by :-

A

`200%`

B

`100%`

C

`800%`

D

`700%`

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The correct Answer is:
To solve the problem, we need to analyze how the inductance of a solenoid changes when its length is halved and its cross-sectional radius is doubled, while keeping the turn density constant. ### Step-by-Step Solution: 1. **Understand the formula for inductance of a solenoid**: The inductance \( L \) of a solenoid is given by the formula: \[ L = \mu_0 n^2 \pi r^2 L \] where: - \( \mu_0 \) is the permeability of free space, - \( n \) is the number of turns per unit length, - \( r \) is the radius of the solenoid, - \( L \) is the length of the solenoid. 2. **Identify the changes in parameters**: - The length of the solenoid is halved: \( L' = \frac{L}{2} \) - The radius of the solenoid is doubled: \( r' = 2r \) 3. **Calculate the new number of turns per unit length**: Since the turn density (number of turns per unit length) is kept constant, \( n \) remains the same. 4. **Substitute the new values into the inductance formula**: The new inductance \( L' \) can be expressed as: \[ L' = \mu_0 n^2 \pi (r')^2 (L') \] Substituting the new values: \[ L' = \mu_0 n^2 \pi (2r)^2 \left(\frac{L}{2}\right) \] Simplifying this: \[ L' = \mu_0 n^2 \pi (4r^2) \left(\frac{L}{2}\right) = \mu_0 n^2 \pi 4r^2 \cdot \frac{L}{2} = 2 \mu_0 n^2 \pi r^2 L \] 5. **Compare the new inductance with the original inductance**: The original inductance \( L \) was: \[ L = \mu_0 n^2 \pi r^2 L \] The new inductance \( L' \) is: \[ L' = 2 \mu_0 n^2 \pi r^2 L \] This shows that: \[ L' = 2L \] 6. **Calculate the increase in inductance**: The increase in inductance is: \[ \Delta L = L' - L = 2L - L = L \] Therefore, the percentage increase in inductance is: \[ \text{Percentage Increase} = \frac{\Delta L}{L} \times 100\% = \frac{L}{L} \times 100\% = 100\% \] ### Final Answer: The inductance of the solenoid increased by **100%**.
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MOTION-ELECTRO MAGNETIC INDUCTION-EXERCISE-2 (Objective problems (Analytical questions))
  1. Two coils are made of copper wires of same length in the first coil t...

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  2. If a current of 2A give rise a magnetic flux of 5xx10^(-5) weber/turn ...

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  3. For a solenoid keeping the turn density constant its length makes halv...

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  4. In the circuit shown in figure bulb will become suddenly bright if :-

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  5. Two bulb of same rating B(1) and B(2) are connected in series with the...

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  6. A constant current I is maintained in a solenoid. Which of lthe follo...

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  7. A coil of copper wire is connected in series with a bulb, a battery an...

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  8. The inductance of a solenoid is 5 henery and its resistance id 5 Omega...

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  9. When a certain circuit consistance of a constant emf . E, an inductanc...

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  10. An LR circuit with a battery is connected at t=0. Which of the follow...

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  11. During 0.1 sec. current in a coil increase from 1A to 1.5 A . If induc...

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  12. A closely wound coil of 100 turns and area of cross-section 1cm^(2) ha...

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  13. In the circuit shown in the figure, what is the value of I(1) just aft...

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  14. Pure inductors each of inductance 3 H are connected as shown in figure...

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  15. The time constant of an inductance coil is 2 xx 10^(-3) s. When a 90 O...

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  16. A toroidal solenoid with an air core has an average radius of 15 cm , ...

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  17. A circular iron, core supports N turns . If a current I produces a mag...

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  18. The self inductance of a toroid is :-

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  19. In Fig (a) and (b), two air-cored solenoids P and Q have been shows. T...

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  20. A small square loop of wire of side l is placed inside a large square ...

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