Circular and Helical Paths of a Charged Particle in a Magnetic Field

“There is only one corner of the universe that you can be certain of improving, and that’s your own self.”

– Aldous Huxley

If a charged particle is projected at right angles to a magnetic field, it follows a circular path of radius ‘r’ which is obtained by equating the magnetic force on it to the centripetal force. Thus we have

 qvB = mv²/r

[qvB is the magnitude of the magnetic force on the particle of charge q moving with speed v at right angles to the magnetic field of flux density B and mv²/r is the centripetal force on the particle of mass m when it moves along a circle of radius r].

From this we get r = mv/qB.

The time taken by the particle to travel once round the circle is the period T of the circular motion.

We have T = 2πr/v = 2πm/qB. The reciprocal of this (qB/2πm) is called the cyclotron frequency.

If the direction of projection of a charged particle is parallel or anti parallel to a magnetic field, it continues to move along its straight line path since there is no magnetic force on it. In general, if a charged particle is projected at an angle θ with respect to a magnetic field, it travels along a helical path because the component (v sinθ) of its velocity perpendicular to the field makes it move along a circle and the component (v cosθ) of velocity parallel to the field makes it move along the field direction. The radius ‘r’ of the helix is obtained by equating the magnetic force to the centripetal force.

Thus we have

 q(vsinθ)B = m(vsinθ)²/r

Therefore, r = m(vsinθ)/qB

The period of the circular component of motion while moving along the helix is the same as that while moving along a pure circle.

Thus period = 2πr/vsinθ = 2πm/qB, on substituting for ‘r’. Note that the time taken by the charged particle to travel once round the circle (in the case of circular path) is the same as the time of travel along one loop of the helix (in the case of helical path) and this is independent of the velocity of the particle.

Pitch of the helical path = Parallel component of velocity× period= vcosθ (2πm/qB).
Now, consider the following questions:

(1) A magnetic field of flux density B along the Y-direction exists in a region of space contained between the planes x = a and x = a + b. The minimum velocity with which a particle of mass ‘m’ and charge ‘q’ should be projected in the X-direction so that it can cross the magnetic field is

(a) qB(a+b)/m

(b) qB/ma

(c) qB/mb

(d) qBa/m

(e) qBb/m

The direction of projection of the particle is perpendicular to the magnetic field. In the magnetic field, the particle moves along a circular path of radius r = mv/qB. On increasing the velocity, the radius of the path increases and just before crossing the field, the particle moves along a semi circle of radius ‘b’ within the field. In this case we have, b = mv/qB so that v = qBb/m. When the velocity exceeds this value, the particle crosses the magnetic field. The correct option therefore is (e).

(2) A proton (mass 1.67×10^–27 kg) is projected with a velocity of 4×10⁶m/s at an angle of 30º with respect to a uniform magnetic field of flux density 0.4 tesla. The path of the proton within the magnetic field is

(a) a circle of radius 2.5cm

(b) a spiral of radius 2.5cm

(c) a spiral of radius 5cm

(d) a spiral of radius 10cm

(e) a circle of radius 5cm.

The path is a spiral (helix) since the direction of projection is not at right angles to the magnetic field.

The radius of the spiral is r = mvsinθ/qB = 1.67×10^–27×4x10⁶×(½) / 1.6×10^–19×0.4 = 0.05 m = 5 cm. So, the correct option is (c). 

You can find some useful questions with solution here.