define the term self inductance of the solenoid obtain the expression for the magnetic energy stored in an inductor of self inductance l build up a current I through it
Answers
Answer:
Self-inductance of the solenoid is the flux associated with the solenoid when unit amount of a current passes through it. Consider a solenoid of self-inductance L, length e, area of cross-section A. Let alternating voltage ε = ε0 sin ωt is applied across it, and back emf induced.
MARK AS BRAINLIEST.... :)
Explanation:
If the current through a coil is altered then the flux through that coil also changes, and this will induce an e.m.f. in the coil itself. This effect is known self-induction and the property of the coil is the self-inductance (L) of the coil, usually abbreviated as the inductance. The self-inductance can be defined in two ways:(a) NF=LI or
(b) Using the equation for the e.m.f. generated: E = - L(dI/dt)
The induced emf is also called back emf . Self-induction is also call inertia of electricity.
Self induction of long solenoid of inductance L
A long solenoid is one which length is very large as compared to its cross section area. the magnetic field inside such a solenoid is constant at any point and given by
B=
l
μ
0
NI
μ
0
=absolutemagneticpermeability
N=totalnumberofturns
Magnetic flux through each turn of solenoid
ϕ=B×areaofeachturn
ϕ=
l
μ
0
NI
×A
totalflux=flux×totalnumberofturns
Nϕ=N(
l
μ
0
NI
×A)..........(1)
If L is the coefficient of inductance of solenoid
Nϕ=LI...............(2)
from equation 1 and 2
LI=N(
l
μ
0
NI
×A)
L=
l
μ
0
N
2
A
(3)
The magnitude of emf is given by
∣e∣ore=L
dt
dI
...............(4)
multiplyingItobothsides
eIdt=LIdI
butI=
dt
dq
Idt=dq
Also work done (dW)= voltage X Charge(dq)
or dW = eXdq = eIdt
substituting the values in equation 4
dW = LIdt
By integrating both sides
∫
0
w
dW=∫
0
I
0
LIdt
W=
2
1
LI
0
2
this work done is in increasing the current flow through inductor is stored as potential energy (U) in the magnetic field of inductor
U=
2
1
LI
0
2