JEE MAIN - Physics (2023 - 13th April Evening Shift - No. 25)
An insulated copper wire of 100 turns is wrapped around a wooden cylindrical core of the cross-sectional area $$24 \mathrm{~cm}^{2}$$. 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 $$1.5 \mathrm{~T}$$ in one direction to $$1.5 ~\mathrm{T}$$ in the opposite direction, the charge flowing through a point in the circuit during the change of magnetic field will be ___________ $$\mathrm{mC}$$.
Answer
60
Explanation
The induced emf in the circuit is given by Faraday's law of electromagnetic induction, which is $\mathcal{E}=-d\phi/dt$, where $\phi$ is the magnetic flux through the circuit.
The magnetic flux through the circuit is proportional to the magnetic field through the core, so we can write $\phi=NBA$, where $N$ is the number of turns in the loop, $B$ is the magnetic field through the core, and $A$ is the cross-sectional area of the core.
As the magnetic field changes from $1.5\mathrm{~T}$ in one direction to $-1.5\mathrm{~T}$ in the opposite direction, the change in magnetic flux is $\Delta\phi=2NBA$.
The induced emf drives a current $I$ through the resistor in the circuit, and the current and the resistance are related by Ohm's law, which is $I=\mathcal{E}/R$. Substituting the expression for $\mathcal{E}$ into this equation, we get $I=-d\phi/dtR$.
The charge $Q$ that flows through the circuit during the change in magnetic field is given by $Q=\int Idt$. Substituting the expression for $I$ into this equation and integrating with respect to time, we get $Q=-\Delta\phi/R$, where $\Delta\phi$ is the change in magnetic flux and $R$ is the resistance of the circuit.
Substituting the given values into this expression, we get:
$$Q=-\frac{2NBA}{R}=-\frac{2(100)(1.5)(24\times10^{-4})}{12}=-0.06\mathrm{~C}=-60\mathrm{~mC}$$
Therefore, the charge flowing through a point in the circuit during the change of magnetic field is $60\mathrm{~mC}$, which is the same as the provided answer.
The magnetic flux through the circuit is proportional to the magnetic field through the core, so we can write $\phi=NBA$, where $N$ is the number of turns in the loop, $B$ is the magnetic field through the core, and $A$ is the cross-sectional area of the core.
As the magnetic field changes from $1.5\mathrm{~T}$ in one direction to $-1.5\mathrm{~T}$ in the opposite direction, the change in magnetic flux is $\Delta\phi=2NBA$.
The induced emf drives a current $I$ through the resistor in the circuit, and the current and the resistance are related by Ohm's law, which is $I=\mathcal{E}/R$. Substituting the expression for $\mathcal{E}$ into this equation, we get $I=-d\phi/dtR$.
The charge $Q$ that flows through the circuit during the change in magnetic field is given by $Q=\int Idt$. Substituting the expression for $I$ into this equation and integrating with respect to time, we get $Q=-\Delta\phi/R$, where $\Delta\phi$ is the change in magnetic flux and $R$ is the resistance of the circuit.
Substituting the given values into this expression, we get:
$$Q=-\frac{2NBA}{R}=-\frac{2(100)(1.5)(24\times10^{-4})}{12}=-0.06\mathrm{~C}=-60\mathrm{~mC}$$
Therefore, the charge flowing through a point in the circuit during the change of magnetic field is $60\mathrm{~mC}$, which is the same as the provided answer.
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