AP EAMCET · PHYSICS · Electromagnetic Induction
A circular coil of radius \(10 \mathrm{~cm}\) and resistance of \(2 \Omega\) is placed with its plane perpendicular to the horizontal component of the earth's magnetic field. It is rotated about its vertical diameter through \(180^{\circ}\) in \(0.25 \mathrm{~s}\). If the magnitude of the induced emf is \(3.8 \times 10^{-3} \mathrm{~V}\), then the number of turns of the coil is
(Horizontal component of earth's magnetic field at the place is \(3 \times 10^{-5} \mathrm{~T}\) )
- A 504 turns
- B 458 turns
- C 302 turns
- D 608 turns
Answer & Solution
Correct Answer
(A) 504 turns
Step-by-step Solution
Detailed explanation
Radius of coil, \(\mathrm{r}=10 \mathrm{~cm}=0.1 \mathrm{~m}\) Resistance of coil, \(R=2 \Omega\) Angle of rotation, \(\theta=180^{\circ}\) Time of rotation, \(\mathrm{T}=0.25 \mathrm{sec}\) Inducided emf, \(\mathrm{E}=3.8 \times 10^{-3} \mathrm{~V}\) Horizontal Magnetic field,…
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