If a coil of 40 turns and area 4.0 , cm^2 is | vedclass

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(B) The induced charge $Q$ flowing through a circuit is given by the formula: $Q = \frac{\Delta \phi}{R}$.Here,$\Delta \phi = N \cdot B \cdot A$,where

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(B) The induced charge $Q$ flowing through a circuit is given by the formula: $Q = \frac{\Delta \phi}{R}$.Here,$\Delta \phi = N \cdot B \cdot A$,where $N$ is the number of turns,$B$ is the magnetic flux density,and $A$ is the area.Given: $N = 40$,$A = 4.0 \, cm^2 = 4.0 	imes 10^{-4} \, m^2$,$Q = 2.0 	imes 10^{-4} \, C$,and $R = 80 \, \Omega$.Substituting these values into the formula: $2.0 	imes 10^{-4} = \frac{40 	imes B 	imes 4.0 	imes 10^{-4}}{80}$.Solving for $B$: $B = \frac{2.0 	imes 10^{-4} 	imes 80}{40 	imes 4.0 	imes 10^{-4}}$.$B = \frac{160 	imes 10^{-4}}{160 	imes 10^{-4}} = 1 \, Wb/m^2$.

If a coil of $40$ turns and area $4.0 \, cm^2$ is suddenly removed from a magnetic field,it is observed that a charge of $2.0 \times 10^{-4} \, C$ flows through the coil. If the resistance of the coil is $80 \, \Omega$,the magnetic flux density in $Wb/m^2$ is:

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