Advertisements
Advertisements
Question
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.
Solution
Given data:
μr = 900
A2 = 4 × 10-4 m2
l = 0.04 m
n1 = 5000
n2 = 20,000
i1 =2A to i2 = 8A
dt = 0.048
M = ?
ε2 = ?
M = μ0 μr n1 n2 A2 l
M = 4π × 10-7 × 900 × 5000 × 20000 × 4 × 10-4 ×0.04
M = 4π × 90 × 4 × 4 × 10-4
M = 18086 × 10-4H
M = 1.81 H
induced emf,
ε2 = M `"di"/"dt"`
ε2 = `1.81 ((8 - 2)/0.04)`
ε2 = `(1.81 xx 6)/(4 xx 106-2)`
= 2.715 × 10-2 V
ε2 = 271.5 V
APPEARS IN
RELATED QUESTIONS
A circular coil with a cross-sectional area of 4 cm2 has 10 turns. It is placed at the centre of a long solenoid that has 15 turns/cm and a cross-sectional area of 10 cm2. The axis of the coil coincides with the axis of the solenoid. What is their mutual inductance?
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?
Assuming that the length of the solenoid is large when compared to its diameter, find the equation for its inductance.
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.
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.