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A 200 turn coil of radius 2 cm is placed...

A 200 turn coil of radius 2 cm is placed coil axially within a long solenoid of 3 cm radius. If the turns density of the solenoid is 90 turns per cm, then calculate mutual inductance of the coil.

Text Solution

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Given : No of turns of small solenoid `N_(2)=200`
Radius of small solenoid `r=2cm=2xx10^(-2)m`
Area of small solenoid `A_(2)=pir^(2)=3.14xx4xx10^(-4)m^(2)`
Turn density of long solenoid `(N_(1))`/(l)=90xx10^(2)m^(*-1)`
To find :
Mutual inductance of the coil M = ?`
Solution :
`M=M_(0)(N_(1))/(l).N_(2)A_(2)`
`=4xx3.14xx10^(-7)xx90xx10^(2)xx200xx3.14xx4xx10^(-4)`
`=2.84xx10^(-3)H.`
`M=2.84MH.`
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