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 need to use the equation for the magnetic force experienced by a charged particle moving in a magnetic field.
The equation for the magnetic force is given by:
F = q * (v × B)
Where:
F is the magnetic force,
q is the charge of the particle,
v is the velocity of the particle, and
B is the magnetic field.
In this case, the force of gravity is balanced by the magnetic force, so we have:
F_gravity = F_magnetic
The force of gravity can be calculated using the equation:
F_gravity = m * g
Where:
m is the mass of the particle, and
g is the acceleration due to gravity, which is approximately 9.8 m/s^2.
We can rewrite the equation as:
m * g = q * (v × B)
Now, let's calculate the force of gravity first:
F_gravity = (2.0x10^-7 kg) * (9.8 m/s^2)
F_gravity = 1.96x10^-6 N
Now, let's rearrange the equation to solve for the magnetic field:
B = (m * g) / (q * v)
Substituting the given values:
B = (2.0x10^-7 kg * 9.8 m/s^2) / (-4.0x10^-5 C * 6.0x10^4 m/s [W])
Calculating this expression will give you the value of the magnetic field B. The direction of the magnetic force is perpendicular to both the velocity v of the particle and the magnetic field B.