A solenoid is 1.5 , m long and its inner diam | vedclass

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(B) The magnetic field $B$ inside a long solenoid is given by $B = \mu_0 n I$,where $n$ is the number of turns per unit length.Total number of turns $

Vedclass · Accepted Answer

$6.31 	imes 10^{-6} \, Wb$

Vedclass · Answer

(B) The magnetic field $B$ inside a long solenoid is given by $B = \mu_0 n I$,where $n$ is the number of turns per unit length.Total number of turns $N = 3 	imes 1000 = 3000$.Length $L = 1.5 \, m$.Number of turns per unit length $n = N/L = 3000 / 1.5 = 2000 \, turns/m$.Current $I = 2.0 \, A$.Radius $r = 2.0 \, cm = 0.02 \, m$.Area of cross-section $A = \pi r^2 = \pi (0.02)^2 = 4\pi 	imes 10^{-4} \, m^2$.Magnetic flux $\phi = B \cdot A = (\mu_0 n I) A$.$\phi = (4\pi 	imes 10^{-7}) 	imes 2000 	imes 2.0 	imes (4\pi 	imes 10^{-4})$.$\phi = 16 \pi^2 	imes 10^{-7} 	imes 4000 	imes 10^{-4} = 64 \pi^2 	imes 10^{-7} 	imes 10^{-1} \approx 64 	imes 9.87 	imes 10^{-8} \approx 6.31 	imes 10^{-6} \, Wb$.

$A$ solenoid is $1.5 \, m$ long and its inner diameter is $4.0 \, cm$. It has three layers of windings of $1000$ turns each and carries a current of $2.0 \, A$. The magnetic flux for a cross-section of the solenoid is nearly:

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