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3 years ago

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A 2.60-m length of wire is held in an east–west direction and moves horizontally to the north with a speed of 0.360 m/s. The Ear

th's magnetic field in this region is of magnitude 50.0 µT and is directed northward and 53.0° below the horizontal. (a) Calculate the magnitude of the induced emf between the ends of the wire.

1 answer:

fenix001 [56]3 years ago

4 0

Answer:

37.4 V

Explanation:

Given:

Length of the wire, l = 2.60m

Velocity of the wire, v = 0.360m/s

Magnetic Field, B = 50.0 µT

Angle θ = 53.0°

The expression for the magnitude of induced emf for a wire is given as;

<em>ε = Blv sinθ </em>-------------(1)

Where;

<em>ε </em>is the induced emf

<em>l</em> is the length of the wire

<em>v</em> is the velocity of the wire

<em>θ</em> is the angle between the magnetic field and the normal of the loop

Substituting the values of l = 2.60m, v = 0.360m/s, B = 50.0 µT and θ = 53.0° into equation (1)

ε = 50 × 2.60 × 0.360 × sin 53.0°

ε = 50 × 2.60 × 0.360 × 0.7986

ε = 37.376 V

ε = 37.4 V

The magnitude of the induced emf between the ends of the wire is 37.4 V

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