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 connected to a resistor. The total resistance in the circuit is 12 ~\Omega . If an externally applied uniform magnetic field in the core along its axis changes from 1.5 ~T in one direction to 1.5 ~T in the opposite direction, the charge flowing through a point in the circuit during the change of magnetic field will be \underline\hspace2cm mC .

Electromagnetic Induction PYQ — JEE Main 2023 · 13 Apr · Shift 2 · Q68 with Solution

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Correct answer: 60

When the magnetic field changes, the total charge that flows is

Q = \frac{1}{R}\int \mathcal{E}\,dt

Using Faraday’s law,

\mathcal{E} = -N\frac{d\Phi}{dt}

So,

Q = \frac{N}{R}\,|\Delta \Phi|

where $\Phi = BA$ since the field is uniform and along the axis of the coil.

Number of turns: $N=100$
Area: $A = 24\,\text{cm}^2 = 24\times 10^{-4}\,\text{m}^2 = 2.4\times 10^{-3}\,\text{m}^2$
Resistance: $R=12\,\Omega$
Magnetic field changes from $+1.5\,\text{T}$ to $-1.5\,\text{T}$

Hence,

\Delta B = (-1.5) - (1.5) = -3.0\,\text{T}

So the magnitude is

|\Delta B| = 3.0\,\text{T}

|\Delta \Phi| = A|\Delta B| = (2.4\times 10^{-3})(3.0)

|\Delta \Phi| = 7.2\times 10^{-3}\,\text{Wb}

Q = \frac{N|\Delta \Phi|}{R} = \frac{100\times 7.2\times 10^{-3}}{12}

Q = \frac{0.72}{12} = 0.06\,\text{C}

0.06\,\text{C} = 60\,\text{mC}

Therefore, the charge flowing through the circuit is

\boxed{60\,\text{mC}}