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(B) Given: Length of solenoid $\ell = 1 \ m$,Diameter $d = 4 \ cm$,Radius $r = 2 \ cm = 0.02 \ m$. Total number of turns $N = 5 	imes 1000 = 5000$. C

Vedclass · Accepted Answer

$4.396 	imes 10^{-2} \ T$

Vedclass · Answer

(B) Given: Length of solenoid $\ell = 1 \ m$,Diameter $d = 4 \ cm$,Radius $r = 2 \ cm = 0.02 \ m$. Total number of turns $N = 5 	imes 1000 = 5000$. Current $I = 7 \ A$. The magnetic field at the centre of a finite solenoid is given by $B = \frac{\mu_0 N I}{2\ell} (\cos 	heta_1 + \cos 	heta_2)$. Since the solenoid is long compared to its radius,we can approximate the field using the formula for an ideal solenoid $B = \mu_0 n I$,where $n = N/\ell$. $B = \frac{\mu_0 N I}{\ell} = \frac{4\pi 	imes 10^{-7} 	imes 5000 	imes 7}{1}$. $B = 4\pi 	imes 10^{-7} 	imes 35000 = 14\pi 	imes 10^{-3} \ T$. $B \approx 14 	imes 3.14159 	imes 10^{-3} \ T = 43.98 	imes 10^{-3} \ T = 4.398 	imes 10^{-2} \ T$. Rounding to the given options,the correct value is $4.396 	imes 10^{-2} \ T$.

$A$ solenoid is $1 \ m$ long and $4 \ cm$ in diameter. It has five layers of windings of $1000$ turns each and carries a current of $7 \ A$. The magnetic field at the centre of the solenoid is

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