The correct option is C x0e2πmv0μ0iq
The situation from the question is as follows:
Let the velocity and acceleration of the charge be →v and →a respectively and the magnetic field be →B.
The force F on the charge = q→v×→B = q(vx→i+vy→j)×μ0i2πx(−→k) = m (ax→i+ay→j)
So,
ax=−qvyμ0i2πxm
ay=qvxμ0i2πxm=dvydt.dxdx = vxdvydx
⇒qμ0i2πxm=dvydx
Now as the magentic field is always normal to the velocity of the charge, the speed of the charge is always v0.
At the instant of maximum seperation between the charge and wire, vx=0 such that vy=v0 directed in the positive y direction and x=x1; and initially vx=v0 directed towrads the wire, vy=0 and x=x0.
So,
∫x1x0qμ0i2πxm.dx=∫v00dvy
Solving the above integral equation we get:
x1=x0e2πmv0μ0iq
Hence, option (c) is correct.