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At a certain location, Earth has a magnetic field of 0.60 ✕ 10−4 T, pointing 75° below the horizontal in a north-south plane. A 15.0 m long straight wire carries a 19 A current. (a) If the current is directed horizontally toward the east, what are the magnitude and direction of the magnetic force on the wire? magnitude N

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lublana

Answer with Explanation:

We are given that

Magnetic field,B=[tex]0.6\times 10^{-4} T[/tex]

[tex]\theta=75^{\circ}[/tex]

Length of wire,l=15 m

Current,I=19 A

a.We have to find the magnitude of magnetic force and direction of magnetic force.

Magnetic force,F=[tex]IBlsin\theta[/tex]

Using the formula

[tex]F=0.6\times 10^{-4}\times 15\times 19sin75[/tex]

[tex]F=16.5\times 10^{-3} N[/tex]

Direction=[tex]tan\theta=cot(90-75)=tan15^{\circ}[/tex]

[tex]\theta=15^{\circ}[/tex]

15 degree above the horizontal  in the northward direction.

Lanuel

The magnitude and direction of the magnetic force on the wire are [tex]16.5 \times 10^{-4}\;Newton[/tex] and 15° respectively.

Given the following data:

  • Magnetic field = [tex]0.60 \times 10^{-4}\;T[/tex]
  • Current = 19 Amperes.
  • Length = 15.0 meters.
  • Angle = 75°.

How to calculate the magnetic force.

Mathematically, the magnitude of a force acting in a magnetic field is given by this formula:

[tex]F=BILsin\theta[/tex]

Where:

  • B is the magnetic field.
  • I is the current.
  • L is the length.
  • [tex]\theta[/tex] is the angle.

Substituting the given parameters into the formula, we have;

[tex]F = 0.60 \times 10^{-4}\times 19 \times 15.0 \times sin 75\\\\F = 16.5 \times 10^{-4}\;Newton[/tex]

For the direction:

[tex]Direction = 90 -75[/tex]

Direction = 15°.

Read more on magnetic field here: https://brainly.com/question/7802337

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