| ⇦ | 
| ⇨ |
Two coaxial solenoids are made by winding thin insulated wire over a pipe of cross-sectional area A=10 cm² and length =20 cm. If one of the solenoids has 300 turns and the other 400 turns, their mutual inductance is (μ₀=4πx10⁻⁷ T mA⁻¹)
Options
(a) 2.4πx10⁻⁴ H
(b) 2.4πx10⁻⁵ H
(c) 4.8πx10⁻⁴ H
(d) 4.8πx10⁻⁵ H
Correct Answer:
2.4πx10⁻⁴ H
Explanation:
No explanation available. Be the first to write the explanation for this question by commenting below.
Related Questions: - Einstein’s photoelectric equation is
- A proton is projected with a speed of 3×10⁶ m/s horizontally from east to west.
- The self induced e.m.f. in a 0.1 H coil when the current in it is changing
- The coefficient of mutual inductance of two coils is 6 mH. If the current flowing
- The potential difference applied to an X-ray tube is 5 kV and the current through it
Topics: Electromagnetic Induction
(76)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- Einstein’s photoelectric equation is
- A proton is projected with a speed of 3×10⁶ m/s horizontally from east to west.
- The self induced e.m.f. in a 0.1 H coil when the current in it is changing
- The coefficient of mutual inductance of two coils is 6 mH. If the current flowing
- The potential difference applied to an X-ray tube is 5 kV and the current through it
Topics: Electromagnetic Induction (76)
Subject: Physics (2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
18000+ students are using NEETLab to improve their score. What about you?
Solve Previous Year MCQs, Mock Tests, Topicwise Practice Tests, Identify Weak Topics, Formula Flash cards and much more is available in NEETLab Android App to improve your NEET score.
Share this page with your friends
Leave a Reply