Question

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

Answer 1

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.