An insulated copper wire of 100 turns is wrapped around a wooden cylindrical core of the cross-sectional area 24 cm^2. The two ends of the wire are…

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

$$\begin{matrix}\epsilon & = & -\frac{d\phi }{dt} \\ & = & iR...(1)\end{matrix}$$ϵ​==​−dtdϕ​iR...(1)​

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

$$\begin{matrix}-\frac{d\phi }{dt} & = & iR \\ & \Rightarrow & -\frac{1}{R}{\int }_{{\phi }_{1}}^{{\phi }_{2}}d\phi =\int idt \\ & \Rightarrow & q=\frac{{\phi }_{1}-{\phi }_{2}}{R}...(2)\end{matrix}$$−dtdϕ​​=⇒⇒​iR−R1​∫ϕ1​ϕ2​​dϕ=∫idtq=Rϕ1​−ϕ2​​...(2)​

Where, ${\phi }_{1},{\phi }_{2}$ϕ1​,ϕ2​ are the magnetic flux associated with the coil from two ends.

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

$$q=\frac{NA}{R}({B}_{1}-{B}_{2})...(3)$$q=RNA​(B1​−B2​)...(3)

where, $N,$N,$A$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.

$$\begin{matrix}q & = & \frac{100\times 24\times {10}^{-4}{m}^{2}\times (1.5T-(-1.5T))}{12\Omega } \\ & = & 0.06C\times \frac{1000\mathrm{mC}}{1C} \\ & = & 60\mathrm{mC}\end{matrix}$$q​===​12Ω100×24×10−4m2×(1.5T−(−1.5T))​0.06C×1C1000mC​60mC​