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.

Net Magnetic Force and Torque on a Current Loop- Two Multiple Choice Questions

If a plane current carrying coil is placed in a uniform magnetic field, generally there will be a torque on the coil, but the net force on the coil will be zero. But, if the current loop is placed in a non-uniform magnetic field, there will be a torque and a net force on the coil. The torque on the loop will be zero only if the plane of the loop is perpendicular to the direction of the magnetic field B. This follows from the expression for torque(τ) given by

τ = n IABsinθ, where ‘n’ is the number of turns, ‘A’ is the area of the loop, ‘I’ is the current and ‘θ’ is the angle between the magnetic field vector B and the area vector A. Note that the area vector is directed normal to the plane of the area. So, if the plane of the current loop is perpendicular to the direction of the magnetic field, the torque is zero. The torque is maximum (equal to nIAB) when the plane of the loop is parallel to the field B.

Now, consider the following MCQ:

A thin copper wire of length ‘L’ is bent to form a single turn plane circular loop and is suspended in a uniform magnetic field ‘B’ which is directed parallel to the plane of the loop. The torque acting on the loop when a current ‘I’ passes in it is 0.1 N. If the same wire were bent to form a plane circular loop of 10 turns, the torque would be

(a) 0.01 N (b) 0.1 N (c) 1 N (d) 0.001 N (e) 10 N

The torque is maximum (equal to nIAB) since the plane of the coil is parallel to the magnetic field. In the first case, since the number of turns (n) is one,

torque, τ = IAB = I×π(L/2π)²×B since the radius of the single turn coil is L/2π.

Therefore, τ = IL²B/4π

When the coil is of ‘n’ turns (with the wire of the same length L),

torque, τ’ = nIA’B = nI×π(L/2πn)²×B = IL²B/4πn.

The torque thus reduces to (1/n)^th of the torque on the single turn coil. Since n = 10, the correct option is 0.01 N.

Let us now consider the following question which is popular among question setters:

A very long straight wire carrying a current ‘I’ is arranged to be coplanar with a rectangular loop of sides ‘a’ and ‘b’ carrying a current ‘i’ as shown. If the nearer parallel side of the loop is at distance ‘r’ from the straight wire, the magnitude of the net magnetic force on the loop is

(a) (μ₀I iab)/[2πr(r+b)]

(b) (μ₀I iab)/[2πr(r+b/2)]

(c) (μ₀I iab)/2πr

(d) (μ₀I iab)/[2π(r+b)]

(e) (μ₀I iab)/[πr(r+b)]

The magnetic forces on the top and bottom sides of the loop are equal and opposite and hence they will get canceled. The force on the left side of the loop is attractive (towards the straight wire) while the force on the right side of the loop is repulsive (away from the straight wire). But they cannot get canceled since the attractive force is greater because of the proximity of the left side to the straight wire.

The magnetic force on the left side = ia(μ₀I/2πr) since (μ₀I/2πr) is the magnetic flux density produced by the straight wire at the distance ‘r’.

Similarly, magnetic force on the right side = ia[μ₀I/2π(r+b)]

Therefore, the net magnetic force on the loop is

ia(μ₀I/2π)[1/r – 1/(r+b)] = (μ₀Iiab)/[2πr(r+b)]

You will find many useful questions (with solution) on magnetic force at apphysicsresources

Two KEAM (Engineering) 2007 Questions from Electrodynamics

The following multiple choice questions appeared in Kerala Engineering Entrance 2007 question paper:

(1) A conducting rod of 1 m length and 1 kg mass is suspended by two vertical wires through its ends. An external magnetic field of 2 T is applied normal to the rod. Now the current to be passed through the rod so as to make the tension in the wires zero is [Take g = 10 m s^–2]

(a) 0.5 A (b) 15 A (c) 5 A (d) 1.5 A (e) 2.5 A

The magnetic force (F) on the rod is given by

F = ILB sinθ where ‘I’ is the current, ‘L’ is the length of the rod, ‘B’ is the magnetic field and ‘θ’ is the angle between the magnetic field and the rod. Since θ = 90º,

F = ILB = I×1×2 = 2I.

The tension in the wires will be zero when the magnetic force balances the weight of the rod.

Therefore, 2I = mg = 1×10 from which I = 5 A.

[It would have been better to mention the magnetic field to be horizontal, in addition to being normal to the rod]

(2) A galvanometer of resistance 20 Ω shows a deflection of 10 divisions when a current of 1 mA is passed through it. If a shunt of 4 Ω is connected and there are 50 divisions on the scale, the range of the galvanometer is

(a) 1 A (b) 3 A (c) 10 mA (d) 30 A (e) 30 mA

This question basically pertains to the conversion of a galvanometer in to an ammeter. Since there are 50 divisions on the scale, the current (I_g) required for full scale deflection is 5 mA.

When the shunt ‘S’ is connected, the maximum current (‘range of galvanometer’, as mentioned in the question) that can be passed into the system (I) is given by

I_gG = (I–I_g)S where ‘G’ is the resistance of the galvanometer.

Therefore, I = I_g(G+S)/S = 5×10^–3(20+4)/4 = 30×10^–3 A = 30 mA.

[You can solve this problem in no time like this: Since the shunt is 1/5^th of the resistance of the galvanometer, it will pass five times the current through the galvanometer. So, the total (maximum) current that can be passed in to the system is six times I_g, which is 6×5 mA = 30 mA]

Magnetic Force on Moving Charge- Kerala Medical Entrance 2007 Question

The following MCQ appeared in Kerala Medical Entrance 2007 question paper:

A proton with energy of 2 MeV enters a uniform magnetic field of 2.5 T normally. The magnetic force on the proton is (Take mass of proton to be 1.6×10^–27 kg)

(a) 3×10^–12 N (b) 8×10^–10 N (c) 8×10^–12 N (d) 2×10^–10 N (e) 3×10^–10 N

The velocity ‘v’ of the proton is to be found first using the expression for kinetic energy E (in joule). Note that E = 2 MeV = 2×10⁶×1.6×10^–19 joule.

We have E = ½ mv² from which v = √(2E/m) = [2×2×10⁶×1.6×10^–19/(1.6×10^–27)]^1/2 = 2×10⁷ ms^–1.

Substituting this value of ‘v’ in the expression for magnetic force (F = qvB), we obtain

F = 1.6×10^–19×2×10⁷×2.5 = 8×10^–12 N.

Note: In the above question we did not take the relativistic increase of mass of the proton into consideration. Since the energy is 2 MeV only and the proton is is fairly heavy, the relativistic increase of mass will be about 0.2% only and you will obtain the velocity of the proton of 2 MeV energy as 1.995×10⁷ ms^–1. So, the magnetic force will be slightly reduced.

You should be aware of the relativistic increase in mass when you deal with questions like the above one, especially if you are preparing for GRE Physics Exam or AP Physics Exam in which you can expect questions of the type given below:

An electron with energy of 2 MeV enters a uniform magnetic field of 2.5 T normally. The magnetic force on the electron is nearly (Take the rest mass of electron to be 9.1×10^–31 kg)

(a) 1.17×10^–10 N (b) 8×10^–10 N (c) 3.35×10^–10 N (d) 8×10^–12 N (e) 3.14×10^–14 N

If you calculate the velocity of the electron without considering the relativistic increase in mass, as we did in the previous question, you will get (nearly) v = 8.39×10⁸ ms^–1 and the magnetic force F = 3.35×10^–10 N, nearly. But, the velocity is more than the velocity of light in free space and therefore is absurd. So, (c) is not the correct option.

If m₀ is the rest mass of the electron and ‘m’ is its mass while moving with kinetic energy E (= 2 MeV), we have

(m – m₀)c² = E, from which m = m₀ + E/c². The energy E is to be substituted in joule in this equation

Therefore, m = 9.1×10^–31 + (2×10⁶ ×1.6×10^–19)/(3 ×10⁸)² = 9.1×10^–31 +3.55×10^–30 = 4.46×10^–30 kg. Note that the mass of the electron has become nearly five times its rest mass.

But m = m₀/√(1 – v²/c²) so that v = c√[1 – (m₀/m)²] = 3×10⁸×√[1 – (9.1×10^–31 /4.46×10^–30)²] = 2.93×10⁸ ms^–1

The magnetic force on the electron is qvB = 1.6×10^–19×2.93×10⁸×2.5 = 1.17×10^–10 N.

Charged Particles in Magnetic and Electric Fields

Here is a multiple choice question which appeared in AIEEE 2002 question paper:
If an electron and a proton having same momenta enter perpendicular to a magnetic field, then
(a) curved path of electron and proton will be same (ignoring the sense of revolution)
(b) they will move undeflected
(c) curved path of electron is more curved than that of the proton
(d) path of proton is more curved
The radius of the circular path of the electron is obtained by equating the magnetic force to the centripetal force: qvB = mv²/r. The radius ‘r’ is therefore given by r = mv/qB. The radius is therefore directly proportional to the momentum (mv) and inversely proportional to the charge (q) of the particle.

The momenta are given as equal in the problem. Since the proton and the electron have the same charge magnitudes and are moving in the same magnetic field B, they will follow paths of the same radius[Option (a)].
You might have noted that the path of a charged particle in an electric field is generally parabolic. This is because of the fact that in a uniform electric field, the electric force on the particle has the same direction everywhere. The motion is similar to the projectile motion in a gravitational field. Now, consider the following question:
A particle of charge ‘+q’ and mass ‘m’ is projected with a velocity ‘v’at an angle ‘θ’ with respect to the horizontal, in an electric field ‘E’ which is directed vertically downwards. If there are no gravitational or magnetic fields, the horizontal range of the particle is
(a) (v²sin2θ)/E (b) (v²sin2θ)/qE (c) (mv²sin2θ)/E (d) (mv²sin2θ)/qE (e) (qmv²sin2θ)/E

In the case of the motion of a projectile in a gravitational field, the expression for horizontal range is R = (v² sin2θ)/g. In the present case, the gravitaional acceleration ‘g’ is replaced by the acceleration produced by the electric field.
Acceleration produced by the electric field = Force/ Mass = qE/m. The correct option therefore is (d).
The following MCQ appeared in AIIMS 2004 question paper:
The cyclotron frequency of an electron gyrating in a magnetic field of 1 T is approximately
(a) 28 MHz (b) 280 MHz (c) 2.8 GHz (d) 28 GHz
The cyclotron frequency is the frequency with which a charged particle describes circular path in a magnetic field and is given by f =qB/2πm with usual notations. [You can get it this way: qvB = mrω² where ω is the angular frequency. Substituting v = ωr in this, we get ω = qB/m. Frquency f = ω/2π =qB/2πm].

Substituting for the mass and charge of the electron, we have

f = (1.6×10^–19 ×1)/ (2π×9.1×10^–31 ) = 28×10⁹ Hz = 28 GHz.

Magnetic Force on Moving Charges

"You may never know what results come of your actions, but if you do nothing, there will be no results."

– Mahatma Gandhi

 
Most of you might be remembering the expression for the magnetic force ‘F’ on a charge ‘q’ moving with velocity ‘v’ in a magnetic field of flux density ‘B’:
F = qvB sinθ where θ is the angle between v and B. Retain the order qvB. It helps you to remember this expression in its vector form:
F = qv×B

The vector form gives you the magnitude (qvB sinθ) of the force and its direction which is along the direction of the cross product vector v×B. You should remember that the direction of v is the direction of motion of a positive charge. If you have a negatively charged particle such as an electron, you should reverse the direction of v to get the direction of the force.
If there is an electric field (E) also in the region, the net force on the charge is given by
F = q(v×B + E).
This is the Lorentz force equation.

Let us consider the following MCQ which appeared in the IIT-JEE Screening 2003 question paper:

For a positively charged particle moving in XY plane initially along the X-axis, there is a sudden change in its path due to the presence of electric and/or magnetic fields beyond P. The curved path is shown in the XY plane and is found to be non-circular.Which one of the following
combinations is possible? (i,j,k are unit
vectors along X,Y,Z directions)
(a) E = 0; B =bi + ck 
(b) E = ai; B =ck + ai
(c) E = 0; B =cj + bk(d) E = ai; B =ck + bj
As the curved path is confined to the XY plane,
the component of magnetic field effective in
deflecting the particle is the Z-component only.
The other component should be the X-component since the X-component cannot deflect the charged particle proceeding along the X-direction. There has to be an electric field since the path of the particle is non-circular. All these conditions are satisfied by option (b).
Now consider the following multiple choice questions:

(1)An electron is projected horizontally from south to north in a uniform horizontal magnetic field acting from west to east. The direction along which it will be deflected by this magnetic field is
(a) vertically upwards 
(b) vertically downwards 
(c) northwards 
(d) southwards 
(e) eastwards
Since the electron is negatively charged, the direction of its velocity is to be reversed for finding the direction of the cross product vector v×B. The correct option therefore is (a). You can use Fleming’s left hand rule (motor rule) also for finding the direction of the magnetic force. But, when you apply the rule, remember that the direction of the conventional current is that of positive charge.
(2) An electron enters a uniform magnetic field of flux density B=2i+6j-√8 k with a velocity V=i+3j-√2 k where i,j and k are unit vectors along the X, Y and Z directions respectively. Then
(a) both speed and path will change 
(b) speed alone will change 
(c) the path will become circular 
(d) the path will become helical (e) neither speed nor path will change
The vectors B and V are parallel as the components of B are twice the components of V. So the magnetic force on the electron is zero and the correct option is (e).
Generally, in problems of the above type, you will have to find the angle between the vectors V and B using cosθ = V.B/ VB. In the present case we have, cosθ = (2+18+4)/ √(12×48) = 1 so that θ= zero. If in a similar problem, the value of θ works out to be 90˚(if cosθ works out to be zero), the correct option would be (c). If the value of θ were neither zero nor 90˚, the path would be helical and the correct option would be (d).

(3) Doubly ionized atoms X and Y of two different elements are accelerated through the same potential difference. On entering a uniform magnetic field, they describe circular paths of radii R₁ and R₂. The masses of X and Y are in the ratio
(a) R₁/R₂ (b) √(R₁/R₂) (c) √(R₂/R₁) (d) (R₂/R₁)² (e) (R₁/R₂)²
Since both atoms are doubly ionized, they have the same charge ‘q’ and since they are accelerated by the same potential difference ‘V’, they have the same kinetic energy, qV. If m₁ and m₂ are the masses and v₁ and v₂ are the velocities of X and Y respectively, we have, ½ m₁v₁² = ½ m₂v₂² from which m₁/m₂ = v₂²/v₁² = [qBR₂/m₂]²/ [qBR₁/m₁]², on substituting for the velocities from the centripetal force equation, mv²/R = qvB.
Therefore, m₁/m₂ = (R₁/R₂)², given by option (e).