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 pla ne 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 answer:
Answer:
Explanation:
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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