A particle with a charge of -4.5 x 10-5 C, a mass of 2.5 x 10-7 kg, and a velocity of 7.0 x 10-5 m/s [E] is traveling 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 field. Ignore any effect due to the Earth’s magnetic field.
To determine the direction and magnitude of the external magnetic field, we need to use the formula for the magnetic force on a moving charged particle:
F = q v B sin(θ)
where:
F is the force experienced by the particle,
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 force of gravity is balanced by the magnetic force, so:
F_gravity = F_magnetic
Since we are looking for the direction of the external field, we can ignore the magnitudes. Thus:
m g = q v B
where:
m is the mass of the particle, and
g is the acceleration due to gravity.
Rearranging the equation:
B = (m g) / (q v)
Now we can substitute the given values:
m = 2.5 x 10^(-7) kg
g = 9.8 m/s^2
q = -4.5 x 10^(-5) C
v = 7.0 x 10^(-5) m/s
B = ((2.5 x 10^(-7) kg) x (9.8 m/s^2)) / ((-4.5 x 10^(-5) C) x (7.0 x 10^(-5) m/s))
Calculating this expression will give us the magnitude of the external magnetic field. Since it is not specified whether the field is in the North or South direction, we can only determine the magnitude.
B = |B|
Now you can plug these values into a calculator or a computer program to find the magnitude of the external magnetic field.