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A solenoid of length 1 m and 0.05 m diam...

A solenoid of length 1 m and 0.05 m diameter has 500 turns. If a current of 2A passes through the coil, calculate (i) the coefficient of self induction of the coil and (ii) the magnetic flux linked with the coil.
Data : `l=1m, d=0.05 m, r=0.025m, N= 5000, I=2A`,
(i) `L=? (ii) phi = ?`

Text Solution

Verified by Experts

(i) `L=(mu_(0)N^(2)A)/(l) = (mu_(0)N^(2)pi r^(2))/(l)`
`= (4 pi xx 10^(-7) xx (5xx10^(2))^(2)xx3.14(0.025)^(2))/(1)`
`=0.616xx10^(-3)`
`:.L=0.616 mH`
(ii) Magnetic flux `phi=LI`
`=0.616xx10^(-3)xx2 = 1.232 xx 10^(-3)`
`phi=1.232` milli weber
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