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A rectangular loop of wire with sides 0.262 and 0.401 m lies in a plane perpendicular to a constant magnetic field (see part a of the drawin
Question
A rectangular loop of wire with sides 0.262 and 0.401 m lies in a plane perpendicular to a constant magnetic field (see part a of the drawing). The magnetic field has a magnitude of 0.764 T and is directed parallel to the normal of the loop’s surface. In a time of 0.153 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.
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Physics
4 years
2021-08-13T00:24:57+00:00
2021-08-13T00:24:57+00:00 2 Answers
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Answers ( )
Answer:
0.525 V
Explanation:
Parameters given:
Sides of rectangular loop = 0.262 m x 0.401 m
Magnetic field, B = 0.764 T
Time, t = 0.153 s
Average EMF induced in a coil is given as:
EMF = (B * N * A) / t
Where N is the number of loops (in this case, 1)
A is the area of the loop = 0.262 * 0.401 = 0.105 m²
EMF = (0.764 * 0.105) / 0.153
EMF = 0.525 V
Answer:
E = 0.262V
Explanation:
Given Area = 0.262m×0.401 m =0.105m², B = 0.764T, Δt = 0.153s
In this time interval the area is halved. This causes the flux to change with time and as a result induce an emf in the loop.
So
ΔФ = BΔA
ΔA = 1/2×0.105m² = –0.0525m²
ΔФ = –0.764×0.0525 = –0.04011Wb
ΔФ/Δt = –0.04011/ 0.153 = –0.262Wb/s
E = –(ΔФ/Δt) = –(–0.262) = 0.262V
E = 0.262V