An electron with a velocity of v = (3 x + 4 y + 2 z) × 106m/s enters a region where there is both a magnetic field and an electric field. The magnetic field is B = (1 x - 2 y + 4 z)T. If the electron experiences no force as it moves, calculate the electric field.
To solve this problem, we can use the formula for the electromagnetic force experienced by a charged particle:
F = q(E + v x B)
where F is the force, q is the charge of the particle, E is the electric field, v is the velocity vector of the particle, and B is the magnetic field.
Given that the electron experiences no force as it moves, we can set the electromagnetic force to zero:
0 = q(E + v x B)
We can start by expanding the cross product term:
0 = q(E + (3 x + 4 y + 2 z) x (1 x - 2 y + 4 z))
Using the properties of cross products, we can calculate each component of the cross product:
(3 x + 4 y + 2 z) x (1 x - 2 y + 4 z) = [(4 x - 8 y + 12 z) - (8 z + 4 x - 2 y) + (2 y - 4 z + 3 x)]
Simplifying, we get:
(3 x + 4 y + 2 z) x (1 x - 2 y + 4 z) = (7 x - 10 y + 6 z)
Substituting this back into the equation, we have:
0 = q(E + (7 x - 10 y + 6 z))
Since this equation holds for all x, y, and z, we can equate the components separately:
0 = 7 q x + E x
0 = -10 q y + E y
0 = 6 q z + E z
Now we have a system of three linear equations. From the given information, we know that B = (1 x - 2 y + 4 z) T, where T represents tesla. Comparing this with the components of E (E x, E y, E z), we can write:
B = (E x, E y, E z)
Substituting the values of B and equating the components:
1 = E x
-2 = E y
4 = E z
Therefore, the electric field is E = (1 x - 2 y + 4 z) N/C (newtons per coulomb).
To summarize, the electric field is given by E = (1 x - 2 y + 4 z) N/C.