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The drawing shows a wire composed of three segments, AB, BC, and CD. There is a current of in the wire. There is also a magnetic field that is the same everywhere and points in the direction of the . The lengths of the wire segments are and . Find the magnitude of the force that acts on each segment.

Respuesta :

Answer:

Explanation:

Given:

Lengths of various segments on the wire:

LAB = 1.3 m

LBC = 0.6 m

LCD = 0.6 m

Current, I = 2.4 A

Magnetc field, B = 0.24 T

Note that this points in the z direction.

F = I × L × B × sinθ

(The length is perpendicular to the magnetic field)

For segment AB,

FAB = I × L × B × sinθ

θ = 90°

= 2.4 × 1.3 × 0.24 × sin 90

= 0.7488 N

= 0.75 N

For segment BC,

FBC = I × L × B × sinθ

θ = 90°

= 2.4 × 0.6 × 0.24 × sin 90

= 0.3456 N

= 0.35 N

For segment CD,

θ = 0° (parallel to the magnetic field)

FBC = I × L × B × sinθ

= 2.4 × 0.6 × 0.24 × sin 0

= 0 N

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