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?

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

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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