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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class12  >>  Electromagnetic Induction
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A square loop of side 10 cm and resistance 0.5 $\Omega $ is placed vertically in the east-west plane. A uniform magnetic field of 0.10 T is set up across the plane in the north-east direction. The magnetic field is decreased to zero in 0.70 s at a steady rate. Determine the magnitudes of induced emf $\varepsilon$ and current $I$ during this time-interval.

(A) $\varepsilon = 1.0mV$ and $I = 2mA$ (B) $\varepsilon = 2.0mV$ and $I = 1mA$ (C) $\varepsilon = 1.2mV$ and $I = 2.1mA$ (D) $\varepsilon = 1.0mV$ and $I = 2.1mA$
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  • 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} $
The angle $\theta$ made by the area vector of the coil with the magnetic field is 45°.
Initial magnetic flux is $\Phi = BA \cos \theta$
$\Phi = \large \frac{(0.1 \times 10^{-2}}{\sqrt 2} $$Wb$
Final Flux $\Phi_{min} = 0$
The change in flux is brought about in 0.70 s.
The magnitude of the induced emf is given by $\varepsilon = \large \frac{|\Delta \Phi_B|}{\Delta T} $
$\varepsilon = \large \frac{|\Delta \Phi_B|}{\Delta T} $$ = \large \frac{|\Delta \Phi - 0|}{\Delta T} $
$\varepsilon = \large\frac{10^{-3}}{\sqrt 2 \times 0.7}$$ = 1.0 mV$
Magnitude of Current $I = \large\frac{\varepsilon}{R}$
Therefore, Magnitude of Current $I = \large\frac{1.0mV}{0.5 \Omega} $$ = 2.0 mA$
answered Mar 10, 2014 by balaji.thirumalai

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