A closely wound solenoid 80; cm long has 5 la | vedclass

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(A) Length of the solenoid,$l = 80\; cm = 0.8\; m$.There are $5$ layers of windings of $400$ turns each on the solenoid.Therefore,the total number of

Vedclass · Accepted Answer

$2.5 	imes 10^{-2}\; T$

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(A) Length of the solenoid,$l = 80\; cm = 0.8\; m$.There are $5$ layers of windings of $400$ turns each on the solenoid.Therefore,the total number of turns on the solenoid,$N = 5 	imes 400 = 2000$.Current carried by the solenoid,$I = 8.0\; A$.The magnitude of the magnetic field inside the solenoid near its centre is given by the formula $B = \frac{\mu_0 N I}{l}$.Here,$\mu_0 = 4\pi 	imes 10^{-7}\; T\; m\; A^{-1}$.Substituting the values: $B = \frac{4\pi 	imes 10^{-7} 	imes 2000 	imes 8.0}{0.8}$.$B = 4\pi 	imes 10^{-7} 	imes 2000 	imes 10 = 8\pi 	imes 10^{-3} \approx 2.512 	imes 10^{-2}\; T$.Thus,the magnitude of the magnetic field is approximately $2.5 	imes 10^{-2}\; T$.

$A$ closely wound solenoid $80\; cm$ long has $5$ layers of windings of $400$ turns each. The diameter of the solenoid is $1.8\; cm$. If the current carried is $8.0\; A$,estimate the magnitude of $B$ inside the solenoid near its centre.

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