JEE Mains · Physics · STD 12 -6. Electromagnetic induction
A circular loop ofradius \(0.3\ cm\) lies parallel to amuch bigger circular loop ofradius \(20\ cm\). The centre of the small loop is on the axis of the bigger loop. The distance between their centres is \(15\ cm\). If a current of \(2.0\ A\) flows through the smaller loop, then the flux linked with bigger loop is
- A \(6.6 \times 10^{-9}\) \( weber\)
- B \(9.1 \times 10^{-11} \) \(weber\)
- C \(6.0 \times 10^{-11} \) \(weber\)
- D \(3.3 \times 10^{-11} \) \(weber\)
Answer & Solution
Correct Answer
(B) \(9.1 \times 10^{-11} \) \(weber\)
Step-by-step Solution
Detailed explanation
Aswe know, Magnetic flux, \(\phi=B . A\) \(\frac{\mu_{0}(2)\left(20 \times 10^{-2}\right)^{2}}{2\left[(0.2)^{2}+(0.15)^{2}\right]} \times \pi\left(0.3 \times 10^{-2}\right)^{2}\) On solving \(=9.216 \times 10^{-11}=9.2 \times 10^{-11}\) \( weber\)
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