Advertisements
Advertisements
प्रश्न
Two air core solenoids have the same length of 80 cm and same cross–sectional area 5 cm2. Find the mutual inductance between them if the number of turns in the first coil is 1200 turns and that in the second coil is 400 turns.
उत्तर
Given data:
l = 80cm
A = 52 = 5 × 10-4 m2
= 80 × 10-2 m
turns = 1500 turns
n1 = `1200/0.8`;
n1 = 1500 turns
n2 = `400/0.8`
n1 = 500 turns
M = ?
Mn = μ0 n1 n2 A2 l
M = 4π × 10-7 × 1500 × 1500 × 5 × 10-4 × 0.8
M = 4π × 10-7 × 75 × 5 × 0.8
M = 3768 × 10-7 H
= 0.3768 × 10-3 H
M = 0.38 mH
APPEARS IN
संबंधित प्रश्न
What do you mean by self-induction?
What is meant by mutual induction?
How will you define the unit of inductance?
Give the physical significance of self-inductance of a coil.
Assuming that the length of the solenoid is large when compared to its diameter, find the equation for its inductance.
An inductor of inductance L carries an electric current i. How much energy is stored while establishing the current in it?
Show that the mutual inductance between a pair of coils is same (M12 = M21).
Determine the self-inductance of 4000 turn air-core solenoid of length 2m and diameter 0.04 m.
A 200 turn circular coil of radius 2 cm is placed co-axially within a long solenoid of 3 cm radius. If the turn density of the solenoid is 90 turns per cm, then calculate mutual inductance of the coil and the solenoid.
The solenoids S1 and S2 are wound on an iron-core of relative permeability 900. The area of their cross-section and their length are the same and are 4 cm2 and 0.04 m, respectively. If the number of turns in S1 is 200 and that in S2 is 800, calculate the mutual inductance between the coils. The current in solenoid 1 is increased from 2A to 8A in 0.04 second. Calculate the induced emf in solenoid 2.