Question

a solenoid with 40 turns per cm carries a current of 1 A.the magnitude energy stored per unit volume is

Answer 1

Given :-

Current = I = 1A

No. Of turns = n = 40/cm = 4000/m

As we know that,

Energy stored = U = B²/2μo

U = (μonI)²/2μo

U = μon²I²/2

Putting the values,

U = (4π × 10-⁷ (4000)² × 1²)/2

U = 2π × 16 × 10-⁷ × 10⁶

U = 32π × 10-¹

U = 3.2π Jm-³.

Hence,

Energy stored per unit volume = U = 3.2π Jm-³.

Answer 2

Given that , A Solenoid with turns per cm carries a current of 1 A [ or , Ampere ] .

Exigency To Find : The magnitude energy stored per unit volume ?

⠀⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━⠀

⠀⠀Given that ,

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⌬⠀Current , I = 1 A [ or Ampere ]

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⌬ ⠀Number of turns per unit length , n = 40 turns per m .

⠀⠀⠀⠀¤ Converting the Number of turns from per m ( metre ) to per cm ( centimeter ) :

$\qquad:\implies \sf Number \:_{\:(\:Turns \: )\:} \:=\: 40 \:turns \:per \: m \:\:\\\\ \qquad:\implies \sf Number \:_{\:(\:Turns \: )\:} \:=\: 40 \times 100 \:turns \:per \: cm \:\:\qquad \because \bigg\lgroup \sf{ 1m \: =\: 100 \:cm}\bigg\rgroup\\\\ \qquad:\implies \underline {\boxed{\pmb{\frak{ Number \:_{\:(\:Turns \: )\:} \:=\: 4000 \:turns \:per \: cm\:}}}} \:\:\bigstar$

⠀⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━⠀

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀¤ Finding Energy stored per unit Volume :

As , We know that ,

⠀⠀⠀⠀⠀⠀⠀ENERGY STORED PER UNIT VOLUME , [ u ] :

$\qquad \star \:\:\underline {\boxed {{\pmb{\sf{  \:u \:=\: \:\bigg\lgroup \sf{ \dfrac{ 1} {2\mu _0 \:} \:\:\times \:B^2}\bigg\rgroup  \:\:Jm^{-3}\:}}}}}$

$\qquad \dashrightarrow \sf \:u \:=\: \:\bigg\lgroup \sf{ \dfrac{ 1} {2\mu _0 \:} \:\:\times \:B^2}\bigg\rgroup  \:\:Jm^{-3}\: \\\\ \qquad \dashrightarrow \sf \:u \:=\: \:\bigg\lgroup \sf{ \dfrac{ 1}{ 2\mu _0 \:} \:\:\times \:\{ \mu_0 n I \}^2 }\bigg\rgroup  \:\:Jm^{-3}\:\\\\ \qquad \dashrightarrow \sf \:u \:=\: \:\bigg\lgroup \sf{ \dfrac{(\:\mu_0 n I\: )^2} {2\mu _0 \:} \: }\bigg\rgroup  \:\:Jm^{-3}\:\\\\ \qquad \dashrightarrow \sf \:u \:=\: \: \dfrac{(\:\mu_0 n I\: )^2} {2\mu _0 \:} \: \:\:Jm^{-3}\:\\\\ \qquad \dashrightarrow \sf \:u \:=\: \: \dfrac{\:\mu_0 n^2 I^2} {2 \:} \: \:\:Jm^{-3}\:\\\\ \qquad \dashrightarrow \sf \:u \:=\: \: \dfrac{\:4 \pi \:\times 10^{-7} \times ( 4000 )^2 \times 1^2} {2 \:} \: \:\:Jm^{-3}\:\\\\ \qquad \dashrightarrow \sf \:u \:=\: \: \:2 \pi \:\times 10^{-7} \times ( 4000 )^2 \times 1\: \:\:Jm^{-3}\:\\\\ \qquad \dashrightarrow \sf \:u \:=\: \: \:32 \pi \:\times 10^{-1} \:\:Jm^{-3}\:\\\\ \qquad \dashrightarrow \sf \:u \:=\: \: \:3.2 \pi \: \:\:Jm^{-3}\:\\\\ \qquad \dashrightarrow \underline {\boxed{\pmb{\frak{ Energy \:Stored \:\:(\:or \;u\:)\:=\:\:3.2 \pi \: \:\:Jm^{-3} \:}}}} \:\:\bigstar$

$\qquad \therefore \underline {\sf Hence, \:Energy \:Stored \:per \:unit \:Volume \:is \:\:\pmb{\bf \:\:3.2 \pi \: \:\:Jm^{-3} }\:.\:}$