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प्रश्न
A 50 cm long solenoid has 400 turns per cm. The diameter of the solenoid is 0.04 m. Find the magnetic flux linked with each turn when it carries a current of 1 A.
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उत्तर
Length of the solenoid, l = 50 cm = 50 × 10-2 m
No. of turns/cm = 400
For 50 cm, No. of turns N = 400 × 50 = 20,000
Diameter of the solenoid = 0.04 m
∴ Radius of the solenoid = 0.02 m
Current passing through the solenoid = 1 A
Area of the solenoid = πr²
= 3.14 × 0.02 × 0.02 m2
Formula :-
Magnetic flux, φ = µ0 n2 AIl
n = `"N"/l`
`therefore φ = (mu_0 "N"^2 "AI")/l`
`φ = (4 xx 3.14 xx 10^-7 xx 20,000 xx 20,000 xx 3.14 xx 0.02 x 0.02 xx 1)/(50 xx 10^-2)`
`= (6310144 xx 10^-7)/(50 xx 10^-2)`
`= 126202.88 xx 10^-7 xx 10^2`
= 126202.88 × 10-5
φ = 1.262 Wb
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