A solenoid of diameter 4cm, length 50cm consists of 2000turns. Calculate its self inductance.
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since , B= ¥Ni/L
CONSIDER,
¥ permeability const.
i Current
N Turns
l length
on substituting values, B= ¥2000 i/50×10^-2 , B=¥40×10^2i
also, flux=L.i here L= SELF INDUCTANCE
Also, flux= B A (area)
A = π 4
finally , flux= BA = ¥4000i
and, flux = Li are equal
therefore,. ¥4000i=Li
hence,
(self INDUCTANCE)L = 4000¥
here,¥= 4π×10^-7
CONSIDER,
¥ permeability const.
i Current
N Turns
l length
on substituting values, B= ¥2000 i/50×10^-2 , B=¥40×10^2i
also, flux=L.i here L= SELF INDUCTANCE
Also, flux= B A (area)
A = π 4
finally , flux= BA = ¥4000i
and, flux = Li are equal
therefore,. ¥4000i=Li
hence,
(self INDUCTANCE)L = 4000¥
here,¥= 4π×10^-7
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