JEE Mains · Physics · STD 12 -6. Electromagnetic induction
An insulated copper wire of \(100\) turns is wrapped around a wooden cylindrical core of the crosssectional 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 \(.........\,mC\).
- A \(50\)
- B \(60\)
- C \(40\)
- D \(30\)
Answer & Solution
Correct Answer
(B) \(60\)
Step-by-step Solution
Detailed explanation
\(\Delta Q =-\frac{\Delta \phi}{ R }=-\left(\frac{\phi_2-\phi_1}{ R }\right)\) \(\phi_1= NBA\) \(\phi_2= NBA\) \(\therefore \Delta Q =\frac{2 NBA }{ R }=\frac{2 \times 100 \times 1.5 \times 24 \times 10^{-4}}{12}\) \(=6 \times 10^{-2}\,C =60\,mC\)
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