An insulated copper wire of 100 turns is wrapped around a wooden cylindrical core of the cross-sectional…

The induced emf $\left(\epsilon \right)$ in the turns of the coil can be written as

$\begin{array}{rcl}\epsilon & =& -\frac{d\phi }{dt}\\ & =& iR ...\left(1\right)\end{array}$

Simplify equation (1) to obtain the required charge through a point in the circuit.

$\begin{array}{rcl}-\frac{d\phi }{dt}& =& iR\\ & \Rightarrow & -\frac{1}{R}{\int }_{{\phi }_1}^{{\phi }_2}d\phi =\int idt\\ & \Rightarrow & q=\frac{{\phi }_1-{\phi }_2}{R} ...\left(2\right)\end{array}$

Where, ${\phi }_1, {\phi }_2$ are the magnetic flux associated with the coil from two ends.

As, $\phi =NBA$, from equation (2), it can be written that

$q=\frac{NA}{R}\left(B_1-B_2\right) ...\left(3\right)$

where, $N,$$A$ are the number of turns and the cross-sectional area of the core respectively.

Substitute the values of the known parameters into equation (3) to calculate the required charge.

$\begin{array}{rcl}q& =& \frac{100\times 24\times {10}^{-4} {\mathrm{m}}^2\times \left(1.5 \mathrm{T}-\left(-1.5 \mathrm{T}\right)\right)}{12 \mathrm{\Omega }}\\ & =& 0.06 \mathrm{C}\times \frac{1000 \mathrm{mC}}{1 \mathrm{C}}\\ & =& 60 \mathrm{mC}\end{array}$