Consider a long solenoid of length l and radi | vedclass

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Question

(A) 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 and $I$ is the current

Vedclass · Accepted Answer

$\mu_0 n^2 \pi r^2 l$

Vedclass · Answer

(A) 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 and $I$ is the current.The magnetic flux $\phi$ through a single turn of the solenoid is $\phi = B 	imes A$, where $A$ is the cross-sectional area of the solenoid.Since the radius is $r$, the area $A = \pi r^2$.Thus, $\phi = (\mu_0 n I) (\pi r^2)$.The total flux $\Phi$ through the solenoid with $N$ total turns is $\Phi = N \phi$.Since $n = N/l$, we have $N = nl$.Therefore, $\Phi = (nl) (\mu_0 n I \pi r^2) = \mu_0 n^2 I \pi r^2 l$.The inductance $L$ is defined as $L = \Phi / I$.Substituting the expression for $\Phi$, we get $L = (\mu_0 n^2 I \pi r^2 l) / I = \mu_0 n^2 \pi r^2 l$.

Consider a long solenoid of length $l$ and radius $r$. If $n$ is the number of turns per unit length and $\mu_0$ is the permeability of free space, the inductance of the solenoid is:

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