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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 $I = 2.1mA$

Toolbox:
• Initial magnetic flux is $\Phi = BA \cos \theta$
• The magnitude of the induced emf is given by $\varepsilon = \large \frac{|\Delta \Phi_B|}{\Delta T}$
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$