A horizontal wire of length 0.78 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 20°, it experiences a magnetic force of 8.3 x 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:
F = BILsinθ
Where:
F is the magnetic force,
B is the magnetic field strength,
I is the current, and
L is the length of the wire.
In this case, we are given:
I = 6.8 A (current)
L = 0.78 m (length)
θ = 20° (angle)
When the wire is horizontal, it experiences no magnetic force, which means the magnetic force is zero. This occurs when the angle θ is 0°.
Using the equation F = BILsinθ, we can substitute the given values to find the magnetic field strength B when the wire is tilted:
8.3 x 10^(-3) N = B * 6.8 A * 0.78 m * sin(20°)
Now, let's solve for B:
B = (8.3 x 10^(-3) N) / (6.8 A * 0.78 m * sin(20°))
B ≈ 0.180 T
Therefore, the magnitude of the external magnetic field is approximately 0.180 Tesla.