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.

Question

Physics

Edith Mckee

Today, 11:38

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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Answers (2)

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Carolyn Caldwell · Answer 1

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

Kathy Moyer · Answer 2

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