A solenoid has N turns,length l,and cross-sec | vedclass

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(C) The magnetic field $B$ at the axial midpoint of a finite solenoid is given by the formula:$B = \frac{\mu_0 n i}{2} (\cos 	heta_1 + \cos 	heta_2)

Vedclass · Accepted Answer

$\frac{N\mu_0 i}{\sqrt{4r^2 + l^2}}$

Vedclass · Answer

(C) The magnetic field $B$ at the axial midpoint of a finite solenoid is given by the formula:$B = \frac{\mu_0 n i}{2} (\cos 	heta_1 + \cos 	heta_2)$For the midpoint,$	heta_1 = 	heta_2 = 	heta$. Thus,$B = \mu_0 n i \cos 	heta$.Here,$n = \frac{N}{l}$ is the number of turns per unit length.From the geometry of the solenoid,$\cos 	heta = \frac{l/2}{\sqrt{r^2 + (l/2)^2}} = \frac{l}{\sqrt{4r^2 + l^2}}$.Substituting these values into the expression for $B$:$B = \mu_0 \left(\frac{N}{l}ight) i \left(\frac{l}{\sqrt{4r^2 + l^2}}ight)$$B = \frac{N \mu_0 i}{\sqrt{4r^2 + l^2}}$

$A$ solenoid has $N$ turns,length $l$,and cross-sectional radius $r$. If a current $i$ flows through the solenoid,what is the magnetic field at the axial midpoint? (Given $l \simeq r$)

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