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29 November, 05:18

A 5-turn square loop (10 cm along aside, resistance = 4.0) is placed in a magnetic field that makes an angle of30o with the plane of the loop. The magnitude of thisfield varies with time according to B = 0.50t2, where t is measured in s andB in T. What is the induced current in the coil att = 4.0 s?

1. 25 mA

2. 50mA

3. 13mA

4. 43mA

5. 5.0 mA

+2
Answers (1)
  1. 29 November, 05:39
    0
    Area A of the coil =.1 x. 1 =.01 m²

    no of turns n = 5

    magnetic field B =.5 t²

    Flux Φ perpendicular to plane passing through it. = nBA sin30

    rate of change of flux

    dΦ/dt = nAdBsin30 / dt

    = nA d/dt (.5t²x. 5)

    = nA x 2 x. 25 x t

    At t = 4s

    dΦ/dt = nA x 2

    = 5x. 01 x 2

    =.1

    current = induced emf / resistance

    =.1 / 4

    =.025 A

    = 25 mA.
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