A charge of -6.17 μC is traveling at a speed of 5.47 × 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 53.2o. A force of magnitude5.37 × 10-3 N acts on the charge. What is the magnitude of the magnetic field?
http://hyperphysics.phy-astr.gsu.edu/hbase/magnetic/movchg.html
F = Q * V x B
F = -6.17*10-6 * 5.57*10^6 * sin 53.2 * B
5.37*10^-3 = | -6.17*10-6 * 5.57*10^6 * sin 53.2 * B |
so
|B| = 5.37*10^-3 / (6.17 * 5.57 * sin 53.2)
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 = |q| * |v| * |B| * sin(θ)
Where:
F is the force experienced by the charged particle
q is the charge of the particle
v is the velocity of the 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 = -6.17 μC (charge of the particle)
v = 5.47 × 10^6 m/s (velocity of the particle)
F = 5.37 × 10^-3 N (force experienced by the particle)
θ = 53.2° (angle between velocity and magnetic field)
First, we need to convert the charge from microcoulombs (μC) to coulombs (C):
1 μC = 1 × 10^-6 C
So, -6.17 μC = -6.17 × 10^-6 C
Now, we can rearrange the formula to solve for the magnetic field (|B|):
|B| = F / (|q| * |v| * sin(θ))
Substituting the given values:
|B| = (5.37 × 10^-3 N) / ((6.17 × 10^-6 C) * (5.47 × 10^6 m/s) * sin(53.2°))
Now, we can calculate the magnitude of the magnetic field:
|B| ≈ 9.82 × 10^-6 T
Therefore, the magnitude of the magnetic field is approximately 9.82 × 10^-6 Tesla.