A solenoid of length 0.4 m and having 500 tu | vedclass

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(C) The magnetic field $B$ inside a long solenoid is given by $B = \mu_0 n I_s$, where $n = \frac{N_s}{L}$.Given: $N_s = 500$, $L = 0.4 \ m$, $I_s = 3

Vedclass · Accepted Answer

$6 	imes 10^{-4} \ Nm$

Vedclass · Answer

(C) The magnetic field $B$ inside a long solenoid is given by $B = \mu_0 n I_s$, where $n = \frac{N_s}{L}$.Given: $N_s = 500$, $L = 0.4 \ m$, $I_s = 3 \ A$.$B = (4\pi 	imes 10^{-7}) 	imes (500 / 0.4) 	imes 3 = 4\pi 	imes 10^{-7} 	imes 1250 	imes 3 = 15000\pi 	imes 10^{-7} = 1.5\pi 	imes 10^{-3} \ T$.The torque $	au$ on a coil in a magnetic field is $	au = N_c I_c A B \sin 	heta$.Given: $N_c = 10$, $I_c = 0.4 \ A$, $r = 0.1 \ m$, $	heta = 90^{\circ}$.Area $A = \pi r^2 = \pi (0.1)^2 = 0.01\pi \ m^2$.$	au = 10 	imes 0.4 	imes (0.01\pi) 	imes (1.5\pi 	imes 10^{-3}) 	imes \sin 90^{\circ}$.$	au = 4 	imes 0.01 	imes 1.5 	imes \pi^2 	imes 10^{-3} 	imes 1$.Using $\pi^2 = 10$:$	au = 0.06 	imes 10 	imes 10^{-3} = 0.6 	imes 10^{-3} = 6 	imes 10^{-4} \ Nm$.

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