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.

Multiple Choice Questions Involving Electromagnetic Induction

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

– Aldous Huxley

Many questions on electromagnetic induction have been discussed on this site earlier. You can access those questions by clicking on the label, ‘electromagnetic induction’ below this post. Today we shall discuss a few more questions in this section.

(1) An object of mass M (Fig.) carrying a charge +Q is released from rest at the top of a smooth inclined plane of inclination θ. A magnetic field of flux density B acts  throughout the region and is directed normally into the plane of the figure. Which one among the following statements is correct?

(a) The acceleration of the object down the incline depends on Q but does not depend on B.

(b) The acceleration of  the object down the incline depends on B but does not depend on Q.

(c) The acceleration of the object down the incline depends on both Q and B.

(d) The acceleration of the object down the incline does not depend on Q and B.

(e) All the above statements are incorrect.

The component of the gravitational force along the incline is Mgsinθ which can produce in the object an acceleration gsinθ down the incline. When the object moves down along the inclined plane, the magnetic force (Lorentz force) acting on the object is directed perpendicular to the inclined plane such that it tries to press the object against the inclined plane.

[Apply Fleming’s left hand rule to arrive at the direction of the magnetic Lorentz force].

Since the magnetic force on the object is perpendicular to the inclined plane, it has no component along the plane. The acceleration of the object down the incline is gsinθ itself and is independent of Q and B [Option (d)].

(2) In the above question suppose the inclined plane is rough and the object is able to slide down. Which one among the following statements is correct?

(a) The acceleration of the object down the incline depends on Q but does not depend on B.

(b) The acceleration of the object down the incline depends on B but does not depend on Q.

(c) The acceleration of the object down the incline depends on both Q and B.

(d) The acceleration of the object down the incline does not depend on Q and B.

(e) All the above statements are incorrect.

In this case the frictional force acting between the object and the inclined plane reduces the acceleration of the object. The frictional force F depends on the normal force N acting on the object.

[Remember that F = μN where μ is the coefficient of friction]

But the normal force (which is Mgcosθ + magnetic Lorentz force) depends on both Q and B since the magnetic Lorentz force is QvB where v is the velocity of the object.

Therefore, the acceleration of the object down the incline depends on both Q and B [Option (c)].

(3) A plane rectangular coil of length 2L and breadth L has N turns of insulated copper wire in it. Initially (at time t = 0) the plane of the coil is perpendicular to a uniform magnetic field of flux density B . The coil rotates with angular velocity ω such that the axis of rotation of the coil is at right angles to the magnetic field (Fig.). If the voltage induced in the coil is V_maxsinωt, the angular velocity of the coil is

(a) V_max / 2NL²B

(b) 2V_max t / NL²B

(c)V_max/NL²B                                                                                            

(d) NL²B/ V_max

(e) 2NL²B/ V_max

The magnetic flux φ linked with the coil at the instant t is given by      

φ = NAB cos ωt where A is the area of the coil.

The voltage V induced in the coil is is given by

            V = – dφ /dt = NABω sin ωt

It is given in the question that V = V_maxsinωt

Therefore, V_max = NABω

This gives ω = V_max /NAB

Since the area A = 2L², we have ω = V_max / 2NL²B

(4) In the transformer shown in the adjoining figure, the secondary winding has taps A, B, C, D and E. The number of turns between taps A and B is 100. The number of turns between taps B and C is 50, the number of turns between C and D is 40 and that between D and E is 10. The primary has 2000 turns and is connected to 220 volt A. C. mains. Between which taps will you obtain 5.5 volt output? 

(a) Between A and B

(b) Between A and B as well as B and E

(c) Between B and C as well as C and E

(d) Between C and D

(e) Between D and E

Since the primary voltage is 220 V and the number of turns in the primary is 2000, the induced voltage per turn is 220/2000 = 0.11 volt.

Therefore, the number of turns required to produce a voltage of 5.5 volt is (5.5)/(0.11) = 50.

Since there are 50 turns between taps B and C as well as between taps C and E, the correct option is (c).

[You can work this out this way also:

Considering the entire secondary of the transformer, the secondary to primary turns ratio is 200/2000 = 1/10. Therefore, the secondary voltage (between the ends A and E) is 220/10 = 22 V. Since the voltage across the entire secondary containing 200 turns is 22 volt, we will get 5.5 volt across 50 turns (between taps B and C as well as between taps C and E)]

 

Lorentz force – An IIT-JEE 2012 Multiple Correct Answer(s) Type Question

“Making the simple complicated is commonplace; making the complicated simple, awesomely simple, that’s creativity.”

– Charles Mingus    

IIT-JEE Multiple Correct Answer(s) Type Questions are designed to have four choices (A), (B), (C) and (D) out of which ONE or MORE are correct. The following question appeared in IIT-JEE 2012 question paper. The question is meant for checking your understanding of Lorentz forces.

Consider the motion of a positive point charge in a region where there are simultaneous uniform electric and magnetic fields E = E₀ ĵ and B = B₀ ĵ. At time t = 0, this charge has velocity v in the x-y plane, making an angle θ with the x-axis. Which of the following option(s) is (are) correct for t > 0?

(A) If θ = 0°, the charge moves in a circular path in the x-z plane.

(B) If θ = 0°, the charge undergoes helical motion with constant pitch along the y-axis.

(C) If θ = 10°, the charge undergoes helical motion with its pitch increasing with time, along the y-axis.

(D) If θ = 90°, the charge undergoes linear but accelerated motion along the y-axis.

Note that the electric and magnetic fields are along the positive y-direction.

If θ = 0, the charged particle is moving at right angles to the electric and magnetic fields (Fig.). The magnetic force (qv×B₀) will make the particle move along a circle and the electric force (qE₀) will push it along the positive y-direction. The path of the particle will therefore be a helix of increasing pitch.

If θ = 10°,the magnetic field will make the particle move along the positive y-direction, following a helical path of constant pitch; but since the electric force (qE₀) pushes it along the positive y-direction, the helical path has increasing pitch.

If θ = 90°, the charged particle is moving parallel to the electric and magnetic fields. The magnetic field has no action on the motion since there is no magnetic force. But the electric force (qE₀) will push the particle along the positive y-direction. The particle is thus in accelerated linear motion. 

Therefore options (C) and (D) are correct.