A magnetic field has a magnitude of 1.2 10-3 T, and an electric field has a magnitude of 5.9 103 N/C. Both fields point in the same direction. A positive 1.8 µC charge moves at a speed of 2.9 106 m/s in a direction that is perpendicular to both fields. Determine the magnitude of the net force that acts on the charge.
Well, the forces are perpendicular, so use resultantforce=sqrt (Fe^2 + Fb^2)
check my thinking. I forever get right hand rules messed up.
To determine the magnitude of the net force acting on the charge, we need to combine the effects of the magnetic field and the electric field on the moving charge. We can use the formula for the net force on a charged particle moving in both an electric and magnetic field:
F = q(E + v x B)
Where:
F is the net force on the charge,
q is the charge of the particle,
E is the electric field,
v is the velocity of the charge, and
B is the magnetic field.
First, let's calculate the cross product v x B:
v x B = (magnitude of v) * (magnitude of B) * sin(theta)
Since the charge is moving perpendicular to both fields, the angle (theta) between v and B is 90°, so sin(theta) = 1. Therefore,
v x B = (magnitude of v) * (magnitude of B)
Now, substitute the given values into the formula:
F = (charge of the particle) * (electric field) + (magnitude of v) * (magnitude of B)
F = (1.8 µC) * (5.9 x 10^3 N/C) + (2.9 x 10^6 m/s) * (1.2 x 10^-3 T)
Now, calculate the product of the charge and the electric field:
(1.8 µC) * (5.9 x 10^3 N/C) = 1.062 x 10^-2 N
And calculate the product of the magnitude of the velocity and the magnetic field:
(2.9 x 10^6 m/s) * (1.2 x 10^-3 T) = 3.48 N
Finally, sum up both parts to find the net force:
F = 1.062 x 10^-2 N + 3.48 N = 3.49 N
Therefore, the magnitude of the net force acting on the charge is 3.49 N.