A horizontal wire of length 0.48 m, carrying a current of 6.8 A, is placed in a uniform external magnetic field. When the wire is horizontal, it experiences no magnetic force. When the wire is tilted upward at an angle of 19° it experiences a magnetic force of 4.00 10-3 N. Determine the magnitude of the external magnetic field.

To determine the magnitude of the external magnetic field, we can use the formula for the magnetic force on a current-carrying wire in a magnetic field:

F = BILsinθ

Where:
- F is the force on the wire,
- B is the magnetic field strength,
- I is the current flowing through the wire,
- L is the length of the wire, and
- θ is the angle between the wire and the magnetic field.

Given:
- The length of the wire, L = 0.48 m
- The current flowing through the wire, I = 6.8 A
- The angle between the wire and the magnetic field, θ = 19°
- The force experienced by the wire, F = 4.00 x 10^-3 N

We can rearrange the formula to solve for the magnetic field strength:

B = F / (ILsinθ)

Substituting the given values:

B = (4.00 x 10^-3 N) / (6.8 A x 0.48 m x sin(19°))

Calculating this expression:

B = (4.00 x 10^-3 N) / (6.8 A x 0.48 m x 0.325)

B ≈ 0.729 T

Therefore, the magnitude of the external magnetic field is approximately 0.729 Tesla.