2.00 m long straight copper wire has a current of 8.0 A in it as it passes through a magnetic field. The magnitude of the force on the conductor is 2.00 N. The angle between the current and the magnetic field is 45°. What is the magnitude of the uniform magnetic field?
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To find the magnitude of the uniform magnetic field, we can use the equation for the magnetic force on a current-carrying wire:
F = BILsinθ
Where:
F is the magnitude of the force on the conductor (given as 2.00 N),
B is the magnitude of the magnetic field (what we want to find),
I is the current in the wire (given as 8.0 A),
L is the length of the wire (given as 2.00 m), and
θ is the angle between the current and the magnetic field (given as 45°).
Rearranging the equation, we can solve for B:
B = F / (ILsinθ)
Substituting the given values, we get:
B = 2.00 N / (8.0 A × 2.00 m × sin(45°))
Now, we can evaluate this expression:
B = 2.00 N / (8.0 A × 2.00 m × sin(45°))
= 2.00 N / (16.0 A·m × sin(45°))
Now, let's calculate the value:
B = 2.00 N / (16.0 A·m × sin(45°))
= 2.00 N / (16.0 A·m × √(2)/2)
= 2.00 N / (16.0 A·m / √2)
Now, simplify the expression further:
B = 2.00 N / (16.0 A·m / √2)
= 2.00 N × ( √2 / 16.0 A·m )
= 0.25 √2 N / (A·m)
Therefore, the magnitude of the uniform magnetic field is approximately 0.25 √2 N / (A·m).