Electromagnetic Induction

JEE Main level, numerical answer, medium

13 of 153

Question 13 of 153

An insulated copper wire of 100100 turns is wrapped around a wooden cylindrical core of the cross-sectional area 24cm224{cm}^{2}. The two ends of the wire are connected to a resistor. The total resistance in the circuit is 12Ω12\Omega. If an externally applied uniform magnetic field in the core along its axis changes from 1.5T1.5T in one direction to 1.5T1.5T in the opposite direction, the charge flowing through a point in the circuit during the change of magnetic field will be _____ mCmC.

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Answer: 60

The induced emf (ϵ)\left(\epsilon \right) in the turns of the coil can be written as
ϵ=−dϕdt=iR...(1)\begin{matrix}\epsilon & = & -\dfrac{d\phi }{dt} \\ & = & iR...\left(1\right)\end{matrix}
Simplify equation (1) to obtain the required charge through a point in the circuit.
−dϕdt=iR⇒−1R∫ϕ1ϕ2dϕ=∫idt⇒q=ϕ1−ϕ2R...(2)\begin{matrix}-\dfrac{d\phi }{dt} & = & iR \\ & \Rightarrow & -\dfrac{1}{R}{\int }_{{\phi }_{1}}^{{\phi }_{2}}d\phi =\int idt \\ & \Rightarrow & q=\dfrac{{\phi }_{1}-{\phi }_{2}}{R}...\left(2\right)\end{matrix}
Where, ϕ1,ϕ2{\phi }_{1},{\phi }_{2} are the magnetic flux associated with the coil from two ends.
As, ϕ=NBA\phi =NBA, from equation (2), it can be written that
q=NAR(B1−B2)...(3)q=\dfrac{NA}{R}\left({B}_{1}-{B}_{2}\right)...\left(3\right)
where, N,N,AA 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.
q=100×24×10−4m2×(1.5T−(−1.5T))12Ω=0.06C×1000mC1C=60mC\begin{matrix}q & = & \dfrac{100\times 24\times {10}^{-4}{m}^{2}\times \left(1.5T-\left(-1.5T\right)\right)}{12\Omega } \\ & = & 0.06C\times \dfrac{1000mC}{1C} \\ & = & 60mC\end{matrix}

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