SATHEE: Electro Magnetic Induction And Alternating Currents Question 494

Electro Magnetic Induction And Alternating Currents Question 494

Question: A wire of fixed lengths is wound on a solenoid of length $ℓ$ and radius r. Its self-inductance is found to be L. Now if same wire is wound on a solenoid of length $ℓ / 2$ and radius r/2, then the self in- ductance will be -

Options:

A) 2L

B) L

C) 4L

D) 8L

Show Answer

Answer:

Correct Answer: A

Solution:

$L = \frac{μ_{0} N^{2} π r^{2}}{ℓ}$ Length of wire=N

$2 π r$ =constant (=C, suppose)

$∴ L = μ_{0} {(\frac{C}{2 π r})}^{2} \frac{π r^{2}}{ℓ}$

$∴ L \propto \frac{1}{ℓ}$

$∴$ Self-inductance will become 2L.

पिछला

अगला

गोपनीयता नीति

द्वारा संचालित Prutor@IITK

sathee@iitk.ac.in

SATHEE का मीडिया कवरेज

खेल स्टोर

ऐप स्टोर

Ask SATHEE

Welcome to SATHEE !
Select from 'Menu' to explore our services, or ask SATHEE to get started. Let's embark on this journey of growth together! 🌐📚🚀🎓

I'm relatively new and can sometimes make mistakes.
If you notice any error, such as an incorrect solution, please use the thumbs down icon to aid my learning.
To begin your journey now, click on "I understand".

Electro Magnetic Induction And Alternating Currents Question 494

Question: A wire of fixed lengths is wound on a solenoid of length ℓ and radius r. Its self-inductance is found to be L. Now if same wire is wound on a solenoid of length ℓ/2 and radius r/2, then the self in- ductance will be -

Options:

Answer:

Solution:

इंजीनियरिंग की तैयारी

चिकित्सा तैयारी

एनसीईआरटी पुस्तकें और समाधान

महत्वपूर्ण संसाधन

अन्य संसाधन

पाठ्यक्रम

परीक्षा मंच

नमूना मॉक टेस्ट

सदस्यता

एनसीईआरटी व्यायाम वीडियो समाधान

Menu

Question: A wire of fixed lengths is wound on a solenoid of length $ℓ$ and radius r. Its self-inductance is found to be L. Now if same wire is wound on a solenoid of length $ℓ / 2$ and radius r/2, then the self in- ductance will be -