AP EAMCET · PHYSICS · Electromagnetic Induction
A uniform magnetic field \(\vec{B}\) is perpendicular to the plane of a circular loop of diameter \(10 \mathrm{~cm}\) formed from wire of diameter \(2 \mathrm{~mm}\) and resistivity \(2 \times 10^{-8} \Omega \mathrm{m}\). If a current of \(11 \mathrm{~A}\) is to be induced in the loop then the rate at which \(\vec{B}\) is to be changed is
- A \(2.8 \mathrm{Ts}^{-}\)
- B \(1.4 \mathrm{~T} \mathrm{~s}^{-1}\)
- C \(3.2 \mathrm{~T} \mathrm{~s}^{-1}\)
- D \(2.4 \mathrm{~T} \mathrm{~s}^{-1}\)
Answer & Solution
Correct Answer
(A) \(2.8 \mathrm{Ts}^{-}\)
Step-by-step Solution
Detailed explanation
\[ \mathrm{i}=\frac{\mathrm{e}}{\mathrm{R}}=\frac{1}{\mathrm{R}} \frac{\mathrm{d} \phi}{\mathrm{dt}}=\frac{\mathrm{A}}{\mathrm{R}} \frac{\mathrm{dB}}{\mathrm{dt}} \] Now, \(A=\) area of loop \(=\pi \times(0.05)^2\) \[ =25 \pi \times 10^{-4} \mathrm{~m}^2 \]…
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