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 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

(1) qv \(\frac{μ_0i}{2a}\)

(2) \(qv\,\frac{μ_0i}{4a}\)

(3) \(qv\frac{μ_0i}{2\pi a}\)

(4) zero

Answer 1

(4) zero

Explanation: When current enters into wire, it gets divided into two equal parts; these meet at a point, diametrically opposite, when leaving the circular wire. Hence the two halves of this wire produce (equal magnitude) magnetic fields at the centre, which act in opposite directions. Therefore the net magnetic field at centre is zero. Hence the particle, while passing through centre, experiences no force; [F = qv x B].