A small coil C with N = 200 turns is mounted | vedclass

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Question

(A) When current $I$ flows through the coil,it experiences a magnetic torque $	au = NIAB \sin 	heta$. Since the magnetic field is perpendicular to t

Vedclass · Accepted Answer

$0.4$

Vedclass · Answer

(A) When current $I$ flows through the coil,it experiences a magnetic torque $	au = NIAB \sin 	heta$. Since the magnetic field is perpendicular to the area vector of the coil,$	heta = 90^\circ$,so $	au = NIAB$.This torque is balanced by the torque due to the additional mass $\Delta m$ placed on the balance pan at distance $l$ from the pivot $O$.The torque due to the weight is $	au_{weight} = \Delta m \cdot g \cdot l$.Equating the two torques: $NIAB = \Delta mgl$.Rearranging for $B$: $B = \frac{\Delta mgl}{NIA}$.Given values: $N = 200$,$I = 22 	imes 10^{-3} \, A$,$A = 1.0 	imes 10^{-4} \, m^2$,$\Delta m = 60 	imes 10^{-6} \, kg$,$l = 0.3 \, m$,$g = 9.8 \, m/s^2$.Substituting the values:$B = \frac{60 	imes 10^{-6} 	imes 9.8 	imes 0.3}{200 	imes 22 	imes 10^{-3} 	imes 1.0 	imes 10^{-4}}$$B = \frac{176.4 	imes 10^{-6}}{4400 	imes 10^{-7}} = \frac{176.4 	imes 10^{-6}}{4.4 	imes 10^{-4}} \approx 0.4 \, T$.

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