A solenoid 60 ;cm long and of radius 4.0; cm | vedclass

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(C) Length of the solenoid,$L = 0.6 \; m$.Radius of the solenoid,$r = 0.04 \; m$.Total number of turns,$N = 3 	imes 300 = 900$.Length of the wire,$l

Vedclass · Accepted Answer

$108$

Vedclass · Answer

(C) Length of the solenoid,$L = 0.6 \; m$.Radius of the solenoid,$r = 0.04 \; m$.Total number of turns,$N = 3 	imes 300 = 900$.Length of the wire,$l = 0.02 \; m$.Mass of the wire,$m = 2.5 	imes 10^{-3} \; kg$.Current in the wire,$i = 6.0 \; A$.Acceleration due to gravity,$g = 9.8 \; m \; s^{-2}$.The magnetic field inside the solenoid is $B = \frac{\mu_0 N I}{L}$,where $I$ is the current in the solenoid.The magnetic force on the wire is $F = B i l = \frac{\mu_0 N I i l}{L}$.For the magnetic force to support the weight of the wire,$F = mg$.Thus,$\frac{\mu_0 N I i l}{L} = mg$.Rearranging for $I$: $I = \frac{mgL}{\mu_0 N i l}$.Substituting the values: $I = \frac{2.5 	imes 10^{-3} 	imes 9.8 	imes 0.6}{4 \pi 	imes 10^{-7} 	imes 900 	imes 6.0 	imes 0.02}$.$I = \frac{0.0147}{1.357 	imes 10^{-4}} \approx 108 \; A$.

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