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
When the current changes from +2A to −2A in 0.05 s, an emf of 8 V is induced in a coil. The co-efficient of self-induction of the coil is
What do you mean by self-induction?
What is meant by mutual induction?
Define self-inductance of a coil interms of
- magnetic flux and
- induced emf
How will you define the unit of inductance?
What do you understand by self-inductance of a coil?
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 long solenoid having 400 turns per cm carries a current 2A. A 100 turn coil of cross-sectional area 4 cm2 is placed co-axially inside the solenoid so that the coil is in the field produced by the solenoid. Find the emf induced in the coil if the current through the solenoid reverses its direction in 0.04 sec.
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.
