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A 1.0-m-long copper wire of diameter 0.10 cm carries a current of 50.0 A to the east. Suppose we apply to this wire a magnetic field that produces on it an upward force exactly equal in magnitude to the wire's weight, causing the wire to "levitate."
A. What is the field's magnitude?
B. What is the field's direction?

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Answer:

1.379 x [tex]10^{-3}[/tex] T. and into the page.

Explanation:

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(a) The magnitude of the magnetic field on the copper wire is 1 x 10⁻⁵ T.

(b) The direction of the field is downwards.

Magnitude of the magnetic field

The magnitude of the magnetic field is calculated as follows;

B = μ₀I/2πL

where;

  • μ₀ is permeability of free space
  • L is length of the wire
  • I is current

B = (4π x 10⁻⁷ x 50) / (2π x 1)

B = 1 x 10⁻⁵ T

Direction of field

The magnetic field will move perpendicular to the direction of the upward force. Thus, the field will point downwards.

Learn more about magnetic field here: https://brainly.com/question/7802337

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