Question

An electric current i enters and leaves a uniform circular wire of radius a through diametrically opposite points. A charged particle q, moving along the axis of the circular wire, passes through its centre at speed v. The magnetic force acting on the particle, when it passes through the centre, has a magnitude equal to
(a) $qv\frac{{\mathrm{\mu }}_{0}i}{2a}$

(b) $qv\frac{{\mathrm{\mu }}_{0}i}{2\mathrm{\pi a}}$

(c) $qv\frac{{\mathrm{\mu }}_{0}i}{a}$

(d) zero

Answer 1

(d) zero

We can use the right-hand thumb rule to get the direction of magnetic field due to the current-carrying wire. Based on this, it can be determined that the direction of magnetic field is along the axis of the wire. Also, the charged particle is moving along the axis. So, no magnetic force will act on it, as the angle between the magnetic field and the velocity of the charged particle may be $0^{°}$ or ${180}^{°}$. So, sin $\theta$ of the angles between velocity and magnetic field is zero.
Also, the force, F=qVB sin $\theta$.
So, the force on the charged particle is zero.