An insulated copper wire of 100 turns is wrapped around a wooden...

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

$\begin{cases}ε & = & - \dfrac{d\phi}{dt} \\ & = & iR . . . \left(1\right)\end{cases}$

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

$\begin{cases}- \dfrac{d\phi}{dt} & = & iR \\ & ⇒ & - \dfrac{1}{R}{∫}_{{\phi}_{1}}^{{\phi}_{2}}d\phi = ∫ idt \\ & ⇒ & q = \dfrac{{\phi}_{1} - {\phi}_{2}}{R} . . . \left(2\right)\end{cases}$

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

$q = \dfrac{NA}{R}\left({B}_{1} - {B}_{2}\right) . . . \left(3\right)$

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.

$\begin{cases}q & = & \dfrac{100 \times 24 \times {10}^{- 4}{m}^{2} \times \left(1 . 5T - \left(- 1 . 5T\right)\right)}{12Ω} \\ & = & 0 . 06C \times \dfrac{1000mC}{1C} \\ & = & 60mC\end{cases}$