The correct option is B vBπE is an integer
Given:
→E=E ^j; →B=B ^j
So, the force due to →E & →B on the charge +q is given by,
→Fe=qE ^j
−→Fm=q[(u^i−v^j)×B ^j]=quB ^k
Now, the acceleration of the charge is given by,
ay=qEm ^j; az=quBm ^k
Assuming particle will return to origin in time t.
Now applying the equation of motion in y− direction for time t
y=uyt+12ayt2
⇒0=−vt+12(qEm)t2
∴t=2vmqE is the time taken by the particle to come to origin due to electric field.
Now due to magnitude field it will move in circular path, time period of its circular motion is given by
T=2πmqB
If the particle completes N number of turns in time of return, then
N×2πmqB=2vmEq
∴N=vBπE
So, here the above value must be an integer so that it complete integral number of rounds.
Hence, option (b) is the correct answer.