A uniform magnetic field of B = 3x10−4 T points in the positive x direction. A charge of 6 μC is fixed at the origin and a charge of −3μC is fixed at (2,0) on the x axis. A charge of 4 μC travels on the y axis at 5.0 x 106 m/s. Neglecting gravity, what force does the moving charge experience when it is 6.0 nm above the origin? Give your answer as a vector (that is, with both magnitude and direction).
To find the force experienced by the moving charge, we can use the formula for the magnetic force on a charged particle:
F = q * v * B * sin(θ)
where:
F is the force on the charge,
q is the charge of the particle,
v is the velocity of the particle,
B is the magnetic field strength, and
θ is the angle between the velocity vector and the magnetic field vector.
In this case, the charge is 4 μC (4 x 10^-6 C), the velocity is given as 5.0 x 10^6 m/s, and the magnetic field strength is 3 x 10^-4 T. We need to determine the angle θ between the velocity and the magnetic field.
Since the moving charge is traveling along the y-axis and the magnetic field is in the positive x direction, the angle θ between the velocity vector and the magnetic field vector is 90 degrees. Thus, sin(θ) = sin(90) = 1.
Plugging in the values into the formula, we have:
F = (4 x 10^-6 C) * (5.0 x 10^6 m/s) * (3 x 10^-4 T) * 1
F = 6 x 10^-3 N
So the magnitude of the force experienced by the moving charge is 6 x 10^-3 N.
Given that the charge is 6.0 nm above the origin along the y-axis, we can treat the force as acting in the negative z direction, perpendicular to the x-y plane. Therefore, the force can be represented as a vector pointing in the negative z-axis direction:
F = 0i + 0j - (6 x 10^-3)k N
So the force experienced by the moving charge is -6 x 10^-3 k N in vector notation, where i, j, and k represent the unit vectors along the x, y, and z axes, respectively.