A positively charged disk is rotated clockwise as shown in Fig. The direction of hte magnetic field at a point A in the plane of the disk is

Explanation

(1) Let.s think it through qualitatively. Initially, the particle is at rest, so the magnetic force on it is zero and the electric field accelerates the charge in the z direction. As it speeds up, a magnetic force develops which, according to , rotates its velocity vector. The faster it goes, the stronger the magnetic field force

becomes , eventually, it curves the particle back around the y axis. At this point the charge is moving against the electrical force. So it begins to slow down. The magnetic force then decreases, and the electrical force takes over, bringing the charge to rest at point P. There the entire proces commences anew, carrying the particle over to point Q, and so on.
(2) This is a case where we need to apply to decide on the path followed by the charge.
Calculate : There being no force in the x direction, the velocity of the particle at any time t can be described by the vector
`vec(v) = v_(y)hat(j) + v_(z)hat(k)`
By applying the equation of Lorentz force, we get
`vec(F) = q[Ehat(k) + (v_(y) hat(j) + v_(z)hat(k)) xx B hat(i)]`
`= q(E - v_(y)B)hat(k) + qv_(z)B hat(j)`
Treating the y and z components separately, we get
`m a_(y) = qv_(z)B`
`m a_(z) = q(E - v_(y)B)`
Let us take
`omega = (qB)/(m)`
Then the equation becomes
`a_(v) = omega v_(z)`
`a_(z) = omega((E)/(B) - v_(y))`.
By differenting the second and using the first to eliminate `a_(y)`, we get
`(d^(2)v_(z))/(d t^(2)) = -omega^(2)v_(z)`. This is the familiar equation of simple harmonic motion. Its solution is given by
`v_(z) = v_(0) sin (omega t + phi)`
At t = 0, `v_(z) = 0`. This means that `phi = 0`. Substituting this expression of `v_(z)` in the second equation, we get
`v_(y) = (E)/(B) - v_(0) cos (omega t)`.
At t = 0, `v_(y)` is also zero. This means that
`v_(0) = (E)/(B)`.
Therefore, we get
`v_(y) (E)/(B)[1- cos (omegat)]`,
`v_(z) = (E)/(B) sin (omega t)`.
By integrating these two expressions, we get
`y = (E)/(omega B)[omega t - sin (omega t)]`
`z = (E)/(omega B)[omega t - cos (omega t)]`
In this expression, if we let
`R = (E)/(omega B)`
and eliminate sine and cosine, we find that
`(y - R omega t)^(2) + (z - R)^(2) = R^(2)`.
This is the equation of a circle of radius R whose center is at `(0, R omega t, R)`. This means that the center travels in y direction at a constant speed of E/B.
The particle moves as though it were a spot on the rim of a wheel, rolling down the y axis at a speed v. This path is called a cycloid.