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, we can use the equation for the magnetic force on a charged particle:
F = qvBsinθ
where F is the force, q is the charge of the particle, v is the velocity, B is the magnetic field strength, and θ is the angle between the velocity and the magnetic field.
In this case, the force of gravity is balanced by the magnetic force, so we can set the magnitude of the two forces equal to each other:
|F_gravity| = |F_magnetic|
Since the charge of the particle is negative, the gravitational force and the magnetic force have opposite directions. The force of gravity is given by:
F_gravity = m*g
where m is the mass of the particle and g is the acceleration due to gravity.
In this case, we can ignore any effect due to the Earth's magnetic field, so the force of gravity is the only external force acting on the particle.
To find the magnitude of the external force, we can set the equations for the gravitational force and the magnetic force equal to each other:
m*g = q*v*B*sinθ
Simplifying the equation, we get:
B = (m*g) / (q*v*sinθ)
Substituting the given values, we have:
B = (2.0x10^-7 kg * 9.8 m/s^2) / (-4.0x10^-5 C * 6.0x10^4 m/s * sin90°)
Since sin90° = 1, the equation becomes:
B = (2.0x10^-7 kg * 9.8 m/s^2) / (-4.0x10^-5 C * 6.0x10^4 m/s)
Calculating the value, we find:
B ≈ -5.1667 T
The negative sign indicates that the external magnetic field is opposite in direction to the gravitational force.
Therefore, the direction of the external force is opposite to the direction of the particle's velocity. The magnitude of the external force is approximately -5.1667 Tesla.