explain the concept of self inductance and calculate the energy stored in coil having coefficient of self inductance 0.3 MS coil is carrying current of 2 ampere
plz give me right ans
nd take 100 points ❤️
Answers
Answer:
Self-inductance of a coil is defined as the ratio of the total flux linked with the coil to the current flowing through it.
L=
I
Nϕ
B
When the current is varied, the flux linked with the coil changes and an e.m.f. is induced in the coil. It is given as
ε=−
dt
d(Nϕ
B
)
=−L
dt
dI
The self-induced e.m.f. is also called back e.m.f. as it opposes any change in current in the circuit.
So, work needs to be done against back e.m.f. in establishing current. This work done is stored as magnetic potential energy. The rate of doing work is given as
dt
dW
=∣ε∣I=LI
dt
dI
(neglecting negative sign)
Thus, the total work done in establishing current from 0 to I is
w=∫dW=
0
∫
I
LIdI=
2
1
LI
2
Self-inductance of a coil is defined as the ratio of the total flux linked with the coil to the current flowing through it.
L= INϕ B
When the current is varied, the flux linked with the coil changes and an e.m.f is induced in the coil. It is given as
ε=− dt d(Nϕ B) =−L dtdI
The self-induced e.m.f is also called back e.m.f. as it opposes any change in current in the circuit So, work needs to be done against back e.m.f. in establishing current This work done is stored as magnetic potential energy. The rate of doing work is given as
dt
dW
=∣ε∣I=LI dtdI
(neglecting negative sign)
Thus, the total work done in establishing current from 0 to I is
W=∫dW 0∫I
LIdI= 1 /2LI 2
pls mark as brainliests answer