An iron pored solenoid having a diameter of 4 cm and a length of 60cm is bound with 4000turns. Find the inductance of the solenoid .what will be the inductance of the solenoid if it has an iron core of relative permiability of 4000?
Answers
Explanation:
.what will be the inductance of the solenoid if it has an iron core of relative permiability of 4000?
Answer:
Self Inductance of long solenoid: Consider a solenoid of length l, cross-section area A. Having total number of turns N. If I current flow through the solenoid there e.m.f. at the centre of solenoid.
B=μ
B=μ o
B=μ o
B=μ o nt
B=μ o ntBut n=
B=μ o ntBut n= 1
B=μ o ntBut n= 1N
B=μ o ntBut n= 1N
B=μ o ntBut n= 1N
B=μ o ntBut n= 1N ⇒B=
B=μ o ntBut n= 1N ⇒B= 1
B=μ o ntBut n= 1N ⇒B= 1μ
B=μ o ntBut n= 1N ⇒B= 1μ o
B=μ o ntBut n= 1N ⇒B= 1μ o
B=μ o ntBut n= 1N ⇒B= 1μ o NI
B=μ o ntBut n= 1N ⇒B= 1μ o NI
B=μ o ntBut n= 1N ⇒B= 1μ o NI
B=μ o ntBut n= 1N ⇒B= 1μ o NI Φ=NBAcosθ
B=μ o ntBut n= 1N ⇒B= 1μ o NI Φ=NBAcosθ⇒L=
B=μ o ntBut n= 1N ⇒B= 1μ o NI Φ=NBAcosθ⇒L= II
B=μ o ntBut n= 1N ⇒B= 1μ o NI Φ=NBAcosθ⇒L= IIμ
B=μ o ntBut n= 1N ⇒B= 1μ o NI Φ=NBAcosθ⇒L= IIμ o
B=μ o ntBut n= 1N ⇒B= 1μ o NI Φ=NBAcosθ⇒L= IIμ o
B=μ o ntBut n= 1N ⇒B= 1μ o NI Φ=NBAcosθ⇒L= IIμ o N
B=μ o ntBut n= 1N ⇒B= 1μ o NI Φ=NBAcosθ⇒L= IIμ o N 2
B=μ o ntBut n= 1N ⇒B= 1μ o NI Φ=NBAcosθ⇒L= IIμ o N 2 IA
B=μ o ntBut n= 1N ⇒B= 1μ o NI Φ=NBAcosθ⇒L= IIμ o N 2 IA
B=μ o ntBut n= 1N ⇒B= 1μ o NI Φ=NBAcosθ⇒L= IIμ o N 2 IA Acosθ
B=μ o ntBut n= 1N ⇒B= 1μ o NI Φ=NBAcosθ⇒L= IIμ o N 2 IA AcosθΦ=
B=μ o ntBut n= 1N ⇒B= 1μ o NI Φ=NBAcosθ⇒L= IIμ o N 2 IA AcosθΦ= l
B=μ o ntBut n= 1N ⇒B= 1μ o NI Φ=NBAcosθ⇒L= IIμ o N 2 IA AcosθΦ= lμ
B=μ o ntBut n= 1N ⇒B= 1μ o NI Φ=NBAcosθ⇒L= IIμ o N 2 IA AcosθΦ= lμ o
B=μ o ntBut n= 1N ⇒B= 1μ o NI Φ=NBAcosθ⇒L= IIμ o N 2 IA AcosθΦ= lμ o
B=μ o ntBut n= 1N ⇒B= 1μ o NI Φ=NBAcosθ⇒L= IIμ o N 2 IA AcosθΦ= lμ o N
B=μ o ntBut n= 1N ⇒B= 1μ o NI Φ=NBAcosθ⇒L= IIμ o N 2 IA AcosθΦ= lμ o N 2
B=μ o ntBut n= 1N ⇒B= 1μ o NI Φ=NBAcosθ⇒L= IIμ o N 2 IA AcosθΦ= lμ o N 2 IA
B=μ o ntBut n= 1N ⇒B= 1μ o NI Φ=NBAcosθ⇒L= IIμ o N 2 IA AcosθΦ= lμ o N 2 IA
B=μ o ntBut n= 1N ⇒B= 1μ o NI Φ=NBAcosθ⇒L= IIμ o N 2 IA AcosθΦ= lμ o N 2 IA
B=μ o ntBut n= 1N ⇒B= 1μ o NI Φ=NBAcosθ⇒L= IIμ o N 2 IA AcosθΦ= lμ o N 2 IA L=
B=μ o ntBut n= 1N ⇒B= 1μ o NI Φ=NBAcosθ⇒L= IIμ o N 2 IA AcosθΦ= lμ o N 2 IA L= l
B=μ o ntBut n= 1N ⇒B= 1μ o NI Φ=NBAcosθ⇒L= IIμ o N 2 IA AcosθΦ= lμ o N 2 IA L= lμ
B=μ o ntBut n= 1N ⇒B= 1μ o NI Φ=NBAcosθ⇒L= IIμ o N 2 IA AcosθΦ= lμ o N 2 IA L= lμ o
B=μ o ntBut n= 1N ⇒B= 1μ o NI Φ=NBAcosθ⇒L= IIμ o N 2 IA AcosθΦ= lμ o N 2 IA L= lμ o
B=μ o ntBut n= 1N ⇒B= 1μ o NI Φ=NBAcosθ⇒L= IIμ o N 2 IA AcosθΦ= lμ o N 2 IA L= lμ o N
B=μ o ntBut n= 1N ⇒B= 1μ o NI Φ=NBAcosθ⇒L= IIμ o N 2 IA AcosθΦ= lμ o N 2 IA L= lμ o N 2
B=μ o ntBut n= 1N ⇒B= 1μ o NI Φ=NBAcosθ⇒L= IIμ o N 2 IA AcosθΦ= lμ o N 2 IA L= lμ o N 2 IA
If the relative permeability of medium inside the solenoid is μ
r
, then,
then,L=
then,L= l
then,L= lμ
then,L= lμ o
then,L= lμ o
then,L= lμ o μ
then,L= lμ o μ r
then,L= lμ o μ r
then,L= lμ o μ r N
then,L= lμ o μ r N 2
then,L= lμ o μ r N 2 IA
It is clear that self inductance of solenoid depends upon the following factors:
(1) On relative permeability of material inside the solenoid: If a soft iron core placed inside the solenoid, the magnetic flux linked with the solenoid increased hence, self inductance of the solenoid will also increases.
(2) On total number of turns in solenoid: If total number of turns in the solenoid increases then self inductance of the solenoid will also increases.
(3) On cross-section area of solenoid: If cross-section area of solenoid increase then self inductance of solenoid will also increases.
(4) On length of solenoid: If length of solenoid increases then its self-inductance decrease.