Sol. The current at time t is I = kt in the outer coil, and 2I = 2kt in the inner one, where k is a constant. Because of these currents the magnetic field in the outer coil is B = μ 0 nkt, whilst in the inner one it is 3B, where n is the number of turns per unit length. The magnetic flux enclosed by the particle’s trajectory of radius r is

Φ = π R 2 × 2B + π r 2 × B = (2R 2 + r 2 ) πμ 0 nkt.

The (constant) magnitude of the induced electric field E can be calculated from the rate of change of magnetic flux with time:

E × = 2 π r = = (2R 2 + r 2 ) πμ 0 nk,

and so

E = .

The charged particle is held in its circular orbit by the magnetic field, and so, from the zero net radial component of the force acting on it, we obtain

= qvB.….. (1)

The particle is accelerated along its circular orbit by the tangential component of the net force according to ma t = qE, where m is the mass and q the electric charge of the particle.

As the magnitude of the electric field is constant, the speed of the particle increases uniformly with time,

v = a t t = t = t.

Inserting this and the value of B into equation (1), we get

t = q μ 0 nk t,

which is satisfied if

= 1, i.e. r = .