A rectangular loop of wire with sides 0.20 and 0.35 $\mathrm{m}$ lies in a plane perpendicular to a constant magnetic field (see part $a$ of the drawing). The magnetic field has a magnitude of 0.65 $\mathrm{T}$ and is directed parallel to the normal of the loop's surface. In a time of 0.18 s, one-half of the loop is then folded back onto the other half, as indicated in part $b$ of the drawing. Determine the magnitude of the average emf induced in the loop.
Added by Victor A.
Step 1
Given: Initial area, $A_{initial} = 0.20 \times 0.35 = 0.07 \, m^2$ Final area, $A_{final} = 0.07 \, m^2$ (since the loop is folded back) Change in area, $\Delta A = A_{final} - A_{initial} = 0.07 - 0.07 = 0 \, m^2$ Show more…
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