A rectangular coil ABCD is hung from one side | vedclass

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(C) In the absence of a magnetic field,the weight added in one pan balances the rectangular coil in the other pan of the balance.Let $M = 500 \, g = 0

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$6$

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(C) In the absence of a magnetic field,the weight added in one pan balances the rectangular coil in the other pan of the balance.Let $M = 500 \, g = 0.5 \, kg$. The balance condition is $Mgl = W_{coil}l$,where $l$ is the arm length of the balance.When a current $I = 9.8 \, A$ is passed through the coil and a magnetic field $B = 0.4 \, T$ is switched on,a magnetic force $F_m = IBL$ acts on the arm $CD$ of length $L = 1.5 \, cm = 1.5 	imes 10^{-2} \, m$.According to Fleming's left-hand rule,the magnetic force acts vertically downwards on the arm $CD$.To regain the balance,an additional mass $m$ must be added to the pan containing the $500 \, g$ mass.The new balance condition is $(M + m)gl = W_{coil}l + F_m l$.Since $Mgl = W_{coil}l$,we have $mgl = F_m l = (IBL)l$.Therefore,$m = \frac{IBL}{g} = \frac{9.8 	imes 0.4 	imes 1.5 	imes 10^{-2}}{9.8}$.$m = 0.4 	imes 1.5 	imes 10^{-2} \, kg = 0.6 	imes 10^{-2} \, kg = 6 	imes 10^{-3} \, kg = 6 \, g$.

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