A charge of -8.9 µC is traveling at a speed of 7.8 106 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 60°. A force of magnitude 4.0 10-3 N acts on the charge. What is the magnitude of the magnetic field?
To find the magnitude of the magnetic field in this scenario, we can use the equation for the magnetic force experienced by a moving charged particle:
F = q * v * B * sin(θ)
Where:
F is the force experienced by the charged particle,
q is the magnitude of the charge,
v is the magnitude of the velocity of the charged particle,
B is the magnitude of the magnetic field,
θ is the angle between the velocity vector and the magnetic field vector.
In this case, we are given:
q = -8.9 µC = -8.9 * 10^-6 C (charge of the particle),
v = 7.8 * 10^6 m/s (speed of the charge),
F = 4.0 * 10^-3 N (force experienced by the charge),
θ = 60° (angle between the velocity and the magnetic field).
We can rearrange the equation to solve for B:
B = F / (q * v * sin(θ))
Plugging in the given values:
B = (4.0 * 10^-3 N) / ((-8.9 * 10^-6 C) * (7.8 * 10^6 m/s) * sin(60°))
Now, to calculate sin(60°), we can use a calculator or reference table to find that sin(60°) = 0.866.
B = (4.0 * 10^-3 N) / ((-8.9 * 10^-6 C) * (7.8 * 10^6 m/s) * 0.866)
Simplifying the equation further:
B = (4.0 * 10^-3 N) / (-6.939 * 10^-11 C⋅m/s)
Finally, performing the division:
B ≈ -5.758 * 10^-8 T
Therefore, the magnitude of the magnetic field is approximately 5.758 * 10^-8 Tesla.