WBJEE · Physics · Electromagnetic Induction
A magnetic field \(B=2 t+4 t^{2}\) (where, \(t_{x}\) time) is applied perpendicular to the plany of a circular wire of radius \(r\) and resistance
R. If all the units are in SI the electric charge that flows through the circular wire during \(t=0 \mathrm{s}\) to \(t=2 \mathrm{s}\) is
- A \(\frac{6 \pi^{2}}{R}\)
- B \(\frac{20 \pi r^{2}}{R}\)
- C \(\frac{32 \pi^{2}}{R}\)
- D \(\frac{48 \pi^{2}}{R}\)
Answer & Solution
Correct Answer
(B) \(\frac{20 \pi r^{2}}{R}\)
Step-by-step Solution
Detailed explanation
Given, \(B=2 t+4 t^{2}\) at \(\quad t=0, B_{1}=0\) and at \(t=2, B_{2}=2 \times 2+4(2)^{2}\) \[ =4+16=20 \mathrm{Wb} / \mathrm{m}^{2} \] We have, \(\Delta Q=\frac{\Delta \phi}{R}=\frac{\pi^{2}\left(B_{2}-B_{1}\right)}{R}\)…
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