A charge of -5.02 μC is traveling at a speed of 8.86 × 106 m/s in a region of space where there is a magnetic field. The angle between the velocity of the charge and the field is 54.6o. A force of magnitude7.85 × 10-3 N acts on the charge. What is the magnitude of the magnetic field?
To find the magnitude of the magnetic field, we can use the formula for the magnetic force experienced by a charged particle moving through a magnetic field:
F = qvBsinθ
Where:
F is the force experienced by the charge,
q is the charge of the particle,
v is the velocity of the particle,
B is the magnetic field, and
θ is the angle between the velocity of the charge and the magnetic field.
In this case, we are given:
q = -5.02 μC = -5.02 × 10^-6 C (convert microcoulombs to coulombs)
v = 8.86 × 10^6 m/s
F = 7.85 × 10^-3 N
θ = 54.6°
We can rearrange the formula to solve for the magnetic field:
B = F / (q * v * sinθ)
Now we can substitute the given values and calculate the magnetic field:
B = (7.85 × 10^-3 N) / ((-5.02 × 10^-6 C) * (8.86 × 10^6 m/s) * sin(54.6°))
B ≈ 0.3 T (Tesla)
Therefore, the magnitude of the magnetic field is approximately 0.3 Tesla.