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Consider a wire of length L =0.30 m that runs north-south on a horizontal surface. There is a current I = 0.50 A flowing north in tie wire. The Earth's magnetic field at this location has a magnitude of 0.50 gauss (or, in SI units. 0.5 x 10^—4 and points north and 38 degrees down from the horizontal, toward the ground. What is the size of the magnetic force on the wire due to the Earth's magnetic field?

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Answer:

The size of the magnetic force on the wire due to the Earth's magnetic field is 4.62 × 10⁻⁶ N.

Step-by-step explanation:

To determine the size of the magnetic force on the wire due to the Earth's magnetic field,

The magnetic force is given by the formula

F = ILB sinθ

Where F is the magnetic force on the wire

I is the electric current in Amperes (A)

L is is the length of wire in meters (m)

B is the magnetic field in Tesla (T)

and θ is the angle between current and magnetic field

From the question,

L = 0.30 m

I = 0.50 A

B = 0.50 gauss = 0.5 × 10⁻⁴ T (NOTE: 1 Gauss = 10⁻⁴ Tesla)

θ = 38°

Now, putting the values into the equation

F = ILB sinθ

F = 0.50 × 0.30 × 0.5 × 10⁻⁴ sin38°

F = 7.5 × 10⁻⁶ (0.61566)

F = 4.62 × 10⁻⁶ N

Hence, the size of the magnetic force on the wire due to the Earth's magnetic field is 4.62 × 10⁻⁶ N.

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