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A 50 cm long solenoid has 400 turns per ...

A 50 cm long solenoid has 400 turns per cm. The diameter of the solenoid is 0.04 m. Find the magnetic flux of a turns when it carries a current of 1 A.

Text Solution

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Geven : Length of the solenoid
`l-=50cm=50xx10^(-2)m`
No. of `"turns"//cm=400`
For 50 cm No. of turns `N=400xx50=20,000`
Diameter of the solenoid = .04 m
Radius of the solenoid `r=(.04)/(2).02m`
To find :
Magnetic flux of a turn `Phi=?`
Current passing through solenoid I = 1 A
Area of the solenoid `A=pii^(2)`
`=3.14xx.02xx.02m^(2)`
Formula :
Magnetic flux `Phi=(mun^(2)AlI)/(l^(2))`
`mu=mu_(0)u_(r)" For air core "mu_(r)=1,mu_(0)=4pixx10^(-7)`m
`n=(N)/(l)`
Solution :
`Phi=(4pi20000xx20000xx3.14xx.02xx.02xx1)/(50xx10^(-2))`
`=(0.06310)/(50)xx10^(3)=1.262Wb.`
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