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Posted by on Sunday, August 12, 2012 at 11:52pm.

A square loop (length along one side = 20 cm) rotates in a constant magnetic field which has a magnitude of 2.0 T. At an instant when the angle between the field and the normal to the plane of the loop is equal to 20 and increasing at the rate of 10/s, what is the magnitude of the induced emf in the loop?

The answer is 4.8 mV, but I keep getting 2.7

I used E = NABwsin(wt)
Because we have no coil, we can take out N so that leave sus with BAwsin(wt)

What am I doing wrong?

THank you

  • Physics - Electromagnetism - , Monday, August 13, 2012 at 1:42am

    ω= Δα/Δt =10º/s=π/18 rad/s.
    A=a²=0.04 m²,
    ωt=20º
    E=- dΦ/dt = - d(B•A•cos(ωt))/dt =
    = B•A• ω•sin(ωt) =
    =2•0.04• (π/18) •sin20º = 4.775•10^-3 V ≈4.8 mV

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