ElectroMagnetic InductionMedium
Question
A wire of fixed length 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 inductance will be:
Options
A.2L
B.L
C.4L
D.8L
Solution
L = 
length of wire = N 2πr = Constant (= C, suppose)
∴ L = m0
∴ L ∝ 
∴ Self inductance will become 2L.
length of wire = N 2πr = Constant (= C, suppose)
∴ L = m0
∴ L ∝ ∴ Self inductance will become 2L.
Create a free account to view solution
View Solution FreeMore ElectroMagnetic Induction Questions
A thin semicircular ring of radius R is falling with its plane vertical in a horizontal magnetic induction . At the posi...A 30 cm long solenoid has 10 turns per cm and area of $5{\text{ }cm}^{2}$. The current through the solenoid coil varies ...AB and CD are fixed conducting smooth rails placed in a vertical plane and joined by a constant current source at its up...A conducting square loop of side L and resistance R moves in its plane with a uniform velocity v perpendicular to one of...Fig. shows a conducting loop being pulled out of a magnetic field with a constant speed v. Which of the four plots shown...