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3 years ago

15

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 o

f 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.

2 answers:

Vesnalui [34]3 years ago

7 0

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

Nesterboy [21]3 years ago

4 0

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

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