The torque required to hold a small circular | vedclass

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(B) The magnetic moment of the coil is $M = N I A$.Given: $N = 10$, $I = \frac{21}{44} \,A$, $A = 1 \,mm^{2} = 10^{-6} \,m^{2}$.$M = 10 	imes \frac{2

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$1.5 	imes 10^{-8} \,N-m$

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(B) The magnetic moment of the coil is $M = N I A$.Given: $N = 10$, $I = \frac{21}{44} \,A$, $A = 1 \,mm^{2} = 10^{-6} \,m^{2}$.$M = 10 	imes \frac{21}{44} 	imes 10^{-6} \,A-m^{2}$.The magnetic field inside the solenoid is $B = \mu_{0} n I_{s}$.Given: $n = 10^{3} \,turns/m$, $I_{s} = 2.5 \,A$, $\mu_{0} = 4\pi 	imes 10^{-7} \,T-m/A$.$B = (4 	imes \frac{22}{7} 	imes 10^{-7}) 	imes 10^{3} 	imes 2.5 \,T$.The torque $	au$ required to hold the coil with its axis perpendicular to the magnetic field is $	au = M B \sin(90^{\circ}) = M B$.$	au = (10 	imes \frac{21}{44} 	imes 10^{-6}) 	imes (4 	imes \frac{22}{7} 	imes 10^{-7} 	imes 10^{3} 	imes 2.5)$.$	au = (10 	imes \frac{21}{44} 	imes 10^{-6}) 	imes (4 	imes \frac{22}{7} 	imes 2.5 	imes 10^{-4})$.$	au = (10 	imes \frac{21}{44} 	imes 10^{-6}) 	imes (22 	imes 10^{-4}) = 1.5 	imes 10^{-8} \,N-m$.

The torque required to hold a small circular coil of $10$ turns, area $1 \,mm^{2}$ and carrying a current of $\left(\frac{21}{44}\right) \,A$ in the middle of a long solenoid of $10^{3} \,turns/m$ carrying a current of $2.5 \,A$, with its axis perpendicular to the axis of the solenoid is

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