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(B) According to Faraday's law of induction,the induced electromotive force $(EMF)$ is given by $\varepsilon = -n \frac{d\phi}{dt}$.By Ohm's law,the i

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$\frac{qR}{nA}$

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(B) According to Faraday's law of induction,the induced electromotive force $(EMF)$ is given by $\varepsilon = -n \frac{d\phi}{dt}$.By Ohm's law,the induced current is $I = \frac{\varepsilon}{R} = -\frac{n}{R} \frac{d\phi}{dt}$.The charge $q$ that flows through the coil is the integral of current over time: $q = \int I dt = \int -\frac{n}{R} \frac{d\phi}{dt} dt = -\frac{n}{R} \int d\phi$.Since the coil is removed from the magnetic field,the change in flux is $\Delta \phi = \phi_{final} - \phi_{initial} = 0 - BA = -BA$.Substituting this into the charge equation: $q = -\frac{n}{R} (-BA) = \frac{nBA}{R}$.Solving for the magnetic flux density $B$,we get $B = \frac{qR}{nA}$.

$A$ coil of $n$ turns and area $A$ is suddenly removed from a magnetic field,and a charge $q$ flows through the coil. If the resistance of the coil is $R$,then the magnetic flux density is (in $Wb/m^2$):

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