A -10 nC charged particle has a velocity v=3i+4j+5k ms-1 as it enters a magnetic field B=10000Tj. Calculate the force vector on the charge. What electric field intensity is required so that the velocity of the charged particle remains constant?
To calculate the force on the charged particle, we can use the equation for the magnetic force on a moving charged particle:
F = q * (v x B)
Where F is the force vector, q is the charge of the particle, v is the velocity vector, and B is the magnetic field vector.
In this case, the charge of the particle is -10 nC, which is equivalent to -10 x 10^-9 C. The velocity vector is given as v = 3i + 4j + 5k m/s, and the magnetic field vector is B = 10000 Tj.
First, let's calculate the vector cross product v x B:
v x B = (3i + 4j + 5k) x (0i + 10000j + 0k)
= (4 * 0 - 5 * 10000)i + (5 * 0 - 3 * 0)j + (3 * 10000 - 4 * 0)k
= -50000i + 0j + 30000k
Now, substitute the values into the equation for the force:
F = (-10 x 10^-9 C) * (-50000i + 0j + 30000k)
= 500 x 10^-6i + 0j - 300 x 10^-6k
So, the force vector on the charged particle is F = 500 x 10^-6i + 0j - 300 x 10^-6k N.
Now, to find the electric field intensity required to keep the velocity of the charged particle constant, we can equate the magnetic force to the electric force:
F = q * E
Where E is the electric field vector.
In this case, the electric field would need to cancel out the magnetic force to keep the velocity constant. Since the magnetic force only has a z-component, the electric field should also have a z-component to balance it out.
Therefore, the electric field intensity required would be E = 300 x 10^-6k V/m.
Note that in real-life scenarios, it is not always possible to cancel out the magnetic force entirely, so this is an idealized case.