A particle with a charge of -4.0x10^-5 C, a mass of 2.0x10^-7 kg and a velocity of 6.0x10^4 m/s [W] is travelling perpendicularly through an external magnetic field such that the force of gravity is balanced by the magnetic force. Determine the direction and magnitude of the external force. Ignore any effect due to the Earth's magnetic field
To determine the direction and magnitude of the external force acting on the particle, we can use the following equation:
F = qvB
Where:
F is the magnitude of the external force
q is the charge of the particle
v is the velocity of the particle
B is the magnitude of the magnetic field
Given values:
q = -4.0x10^-5 C
v = 6.0x10^4 m/s [W]
B is unknown
To find B, we can use the fact that the force of gravity is balanced by the magnetic force. Since we know the mass of the particle, we can equate the gravitational force and the magnetic force:
mg = qvB
Where:
m is the mass of the particle
g is the acceleration due to gravity
Given values:
m = 2.0x10^-7 kg
g is the acceleration due to gravity, which is approximately 9.8 m/s^2
Rearranging the equation, we can solve for B:
B = (mg)/(qv)
Substituting the values:
B = (2.0x10^-7 kg * 9.8 m/s^2) / (-4.0x10^-5 C * 6.0x10^4 m/s [W])
Calculating this, we get:
B ≈ -3.3x10^-3 T
Now that we know the magnetic field magnitude, we can substitute this value back into the original equation to find the external force:
F = qvB
Substituting the values:
F = (-4.0x10^-5 C) * (6.0x10^4 m/s [W]) * (-3.3x10^-3 T)
Calculating this, we get:
F ≈ 7.9x10^-1 N
Therefore, the magnitude of the external force acting on the particle is approximately 7.9x10^-1 N, and the direction is perpendicular to both the velocity of the particle and the magnetic field.