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www.GOALias.blogspot.comElectromagneticInductionSolution The angle θ made by the area vector of the coil with themagnetic field is 45 ° . From Eq. (6.1), the initial magnetic flux isΦ = BA cos θ=–20.1 × 10 Wb2Final flux, Φ min= 0The change in flux is brought about in 0.70 s. From Eq. (6.3), themagnitude of the induced emf is given byεΔΦΔtB= =( Φ – 0)Δt–310= 1.0mV2 × 0.7 =And the magnitude of the current is–310 VI = ε2mAR= 0.5Ω=Note that the earth’s magnetic field also produces a flux through theloop. But it is a steady field (which does not change within the timespan of the experiment) and hence does not induce any emf.EXAMPLE 6.2Example 6.3A circular coil of radius 10 cm, 500 turns and resistance 2 Ω is placedwith its plane perpendicular to the horizontal component of the earth’smagnetic field. It is rotated about its vertical diameter through 180°in 0.25 s. Estimate the magnitudes of the emf and current induced inthe coil. Horizontal component of the earth’s magnetic field at theplace is 3.0 × 10 –5 T.SolutionInitial flux through the coil,Φ B (initial)= BA cos θ= 3.0 × 10 –5 × (π ×10 –2 ) × cos 0º= 3π × 10 –7 WbFinal flux after the rotation,Φ B (final)= 3.0 × 10 –5 × (π ×10 –2 ) × cos 180°= –3π × 10 –7 WbTherefore, estimated value of the induced emf is,ΔΦε = NΔ t= 500 × (6π × 10 –7 )/0.25= 3.8 × 10 –3 VI = ε/R = 1.9 × 10 –3 ANote that the magnitudes of ε and I are the estimated values. Theirinstantaneous values are different and depend upon the speed ofrotation at the particular instant.EXAMPLE 6.3209