If a coil of 40 turns and area 4 , cm^2 is su | vedclass

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(B) Given: Number of turns $N = 40$,Area $A = 4 \, cm^2 = 4 	imes 10^{-4} \, m^2$,Charge $q = 2 	imes 10^{-4} \, C$,Resistance $R = 80 \, \Omega$.Th

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(B) Given: Number of turns $N = 40$,Area $A = 4 \, cm^2 = 4 	imes 10^{-4} \, m^2$,Charge $q = 2 	imes 10^{-4} \, C$,Resistance $R = 80 \, \Omega$.The induced charge $q$ is given by the formula $q = \frac{\Delta \phi}{R}$,where $\Delta \phi$ is the change in magnetic flux.Since the coil is removed from the magnetic field,the final flux is $0$. Thus,$\Delta \phi = N \cdot A \cdot B$.Substituting the values: $q = \frac{N \cdot A \cdot B}{R}$.Rearranging for $B$: $B = \frac{q \cdot R}{N \cdot A}$.$B = \frac{(2 	imes 10^{-4} \, C) 	imes (80 \, \Omega)}{40 	imes (4 	imes 10^{-4} \, m^2)}$.$B = \frac{160 	imes 10^{-4}}{160 	imes 10^{-4}} = 1 \, Wb \, m^{-2}$.

If a coil of $40$ turns and area $4 \, cm^2$ is suddenly removed from a magnetic field,it is observed that a charge of $2 \times 10^{-4} \, C$ 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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