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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class12  >>  Electromagnetic Induction
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A circular coil of radius $10\; cm, 500$ turns and resistance $2 \;\Omega$ 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 $0.25\; s$. Estimate the magnitudes of the emf and current induced in the coil. Horizontal component of the earth’s magnetic field at the place is $3.0 × 10^{–5}$ T.

(A) $\varepsilon = 1.9 \times 10^{-3}$ and $I = 1.9 \times 10^{-3}$ (B) $\varepsilon = 3.8 \times 10^{-3}$ and $I = 1.9 \times 10^{-3}$ (C) $\varepsilon = 1.9 \times 10^{-3}$ and $I = 3.8 \times 10^{-3}$ (D) $\varepsilon = 0.9 \times 10^{-3}$ and $I = 1.8 \times 10^{-3}$
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1 Answer

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Toolbox:
  • Initial magnetic flux is $\Phi = BA \cos \theta$
  • The magnitude of the induced emf is given by $\varepsilon = \large \frac{|\Delta \Phi_B|}{\Delta T} $
  • Magnitude of Current $I = \large\frac{\varepsilon}{R}$
Initial magnetic flux is $\Phi_{Initial} = BA \cos \theta$
$\Phi_{Initial} = 3.0 \times 10^{-5} \times (\pi \times 10^{-2}) \cos 0^{\circ}$
$\Phi_{Initial} = 3.0\;\pi \times 10^{-7} Wb$
Final magnetic flux $\Phi_{Final} = 3.0 \times 10^{-5} \times (\pi \times 10^{-2}) \cos 0^{\circ}$
The magnitude of the induced emf is given by $\varepsilon = N \large \frac{\Delta \Phi} {\Delta t} $
$\Rightarrow$ Estimated value of induced emf $ = \varepsilon = N \large \frac{\Delta \Phi} {\Delta t} $
$\varepsilon =500 \large\frac{ 6\pi \times 10^{-7}}{0.25}$$=3.8 \times 10^{-3} V$
Magnitude of Current $I = \large\frac{\varepsilon}{R}$
$I = 1.9 \times 10^{-3} A$
answered Mar 10, 2014 by balaji.thirumalai
edited Mar 12, 2014 by balaji.thirumalai
 

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