JEE Mains Physics · Moving Charges and Magnetism

Circular and Helical Paths of Charges

A charge moving across a uniform field goes round a circle of radius r = mv/qB with period 2πm/qB, the same at every speed; a velocity at an angle to B makes the circle a helix.

All Moving Charges and Magnetism notes JEE Mains Physics strategy

Why this matters

Twenty-six PYQs, nineteen of them multiple choice, and one from 2026. Thirteen compare the radii of two or three particles, protons, deuterons, alpha particles or ions, at equal speed, momentum, kinetic energy or accelerating voltage. Five find a radius, a mass, a distance or a graph outright. Eight use the period: three are helices, two ask for the period or the frequency of revolution and three are about the cyclotron.

Concept 1 of 3: Radius of a charge's circular path and how it compares between particles

The magnetic force qvB supplies the centripetal force mv²/r, so r = mv/qB: the radius is the momentum divided by qB. Every comparison question is about which quantity the particles share. Write r in terms of that shared quantity, and the ratio is read off from the masses and charges that are left.

Definition

$qvB = \dfrac{mv^{2}}{r}$ , so $r = \dfrac{mv}{qB} = \dfrac{p}{qB} = \dfrac{\sqrt{2mK}}{qB} = \dfrac{1}{B}\sqrt{\dfrac{2mV}{q}}$ (accelerated from rest through V).
Same momentum: $r \propto 1/q$ . Same speed: $r \propto m/q$ . Same kinetic energy: $r \propto \sqrt{m}/q$ . Same accelerating voltage: $r \propto \sqrt{m/q}$ .
Proton (m, e), deuteron (2m, e), alpha particle (4m, 2e). An alpha has the same m/q as a deuteron.
Curvature is $1/r$ : a larger radius is a SMALLER curvature, and a particle with the smaller radius is deflected more.
The magnetic force on each is $qvB$ ; at equal momentum v is $p/m$ .

Radius of the circle

r = \frac{mv}{qB} = \frac{p}{qB} = \frac{\sqrt{2mK}}{qB} = \frac{1}{B}\sqrt{\frac{2mV}{q}}

Worked example

A proton, a deuteron and an alpha particle are each accelerated from rest through the same potential difference and enter the same uniform field at right angles. Find the ratio of the radii of their paths.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2022 · 14 June 2022 · Q99Moderate

Example 1 · Moving Charges and Magnetism · Circular and Helical Paths of Charges

A proton, a deuteron and an

\alpha

-particle with same kinetic energy enter into a uniform magnetic field at right angle to magnetic field. The ratio of the radii of their respective circular paths is:

Equal energy and equal voltage are different conditions

At equal kinetic energy r goes as √m/q; after the same accelerating voltage r goes as √(m/q). For an alpha particle against a proton the two give different ratios, so read which one the question fixes.

Curvature is not radius

Curvature is 1/r. A particle with a larger radius has a smaller curvature and is deflected less. Reading 'smaller curvature' as 'smaller radius' flips the comparison.

Radius goes as the square root of kinetic energy

r = √(2mK)/qB, so doubling K multiplies r by √2, not 2. The radius is proportional to momentum, not to energy.

Drill 12 more on radius of a charge's circular path and how it compares between particles

Concept 2 of 3: Finding a radius, a mass or a field from r = mv/qB

The same equation, r = mv/qB, works the other way round: give it the radius and it returns the mass, the speed or the field. The care is all in the units: charge in coulombs, mass in kilograms, energy in joules, and the radius rather than the diameter of a semicircle.

Definition

$r = \dfrac{mv}{qB}$ ; rearranged, $m = \dfrac{qBr}{v}$ or, after acceleration through V, $m = \dfrac{qB^{2}r^{2}}{2V}$ .
Energy in eV: multiply by $1.6 \times 10^{-19}$ to get joules. Mass in u: multiply by $1.66 \times 10^{-27}$ kg.
A charge that enters a field region and comes back out has gone round a semicircle; the distance between entry and exit is the diameter, 2r.
r is proportional to p, so a graph of r against momentum is a straight line through the origin; against kinetic energy it grows as $\sqrt K$ .
Two regions with different fields: the charge goes round a semicircle of diameter $2mv/qB$ in each; the shift per cycle is the difference of the two diameters.
A quantised orbit in a magnetic field: set the angular momentum $mvr = nh/2\pi$ and use $mv = eBr$ from the circle, then solve for r.

Mass from a measured radius

m = \frac{qBr}{v} = \frac{qB^{2}r^{2}}{2V}

Worked example

An alpha particle (mass

6.4 \times 10^{-27}

kg, charge

3.2 \times 10^{-19}

C) moves at

2 \times 10^{6}

m/s at right angles to a field of 0.5 T. Find the radius of its path.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2023 · 1 February 2023 · Q24Moderate

Example 2 · Moving Charges and Magnetism · Circular and Helical Paths of Charges

A charge particle of

2\mu C

accelerated by a potential difference of

100\text{ }V

enters a region of uniform magnetic field of magnitude

4mT

at right angle to the direction of field. The charge particle completes semicircle of radius

3\text{ }cm

inside magnetic field. The mass of the charge particle is____

\times10^{- 18}\text{ }kg

Energy in eV must be turned into joules

Put kinetic energy into √(2mK) in joules: 1 eV = 1.6 × 10⁻¹⁹ J. A keV is a thousand of those.

Radius, not diameter

When a charge goes round half a circle and exits, the gap between entry and exit is 2r. Putting that gap into r = mv/qB doubles the mass or halves the field.

Singly ionised means charge e

A singly ionised atom has q = e whatever its mass number; the mass number gives only the mass, in u.

Drill 4 more on finding a radius, a mass or a field from r = mv/qb

Concept 3 of 3: Period of revolution, the helix and the cyclotron

A faster charge goes round a bigger circle, and the two effects cancel exactly: the time for one turn, 2πm/qB, does not depend on the speed. That is what makes a cyclotron work, since an alternating voltage of fixed frequency stays in step as the particle speeds up. If the velocity has a part along B, that part is untouched, and the circle is drawn out into a helix.

Definition

Period $T = \dfrac{2\pi m}{qB}$ , frequency $f = \dfrac{qB}{2\pi m}$ , angular frequency $\omega = \dfrac{qB}{m}$ : all independent of speed and radius.
Velocity at angle $\theta$ to B: $v_\perp = v\sin\theta$ makes the circle, $r = \dfrac{mv_\perp}{qB}$ ; $v_\parallel = v\cos\theta$ carries it along B.
Pitch (distance along B per turn): $p = v_\parallel T = \dfrac{2\pi m v\cos\theta}{qB}$ . In n turns it moves $n$ pitches.
v along B: a straight line. v at right angles to B: a circle. Anything in between: a helix with its axis along B.
Cyclotron: the oscillator frequency equals $qB/2\pi m$ . The largest kinetic energy, at the dee radius R, is $K_{\max} = \dfrac{q^{2}B^{2}R^{2}}{2m}$ .
The particle crosses the gap twice per revolution, gaining qV each time, so the number of revolutions to reach K is $\dfrac{K}{2qV}$ .

Period, pitch and cyclotron energy

T = \frac{2\pi m}{qB} \qquad p = v\cos\theta\,T \qquad K_{\max} = \frac{q^{2}B^{2}R^{2}}{2m}

Worked example

A particle of mass

10^{-6}

kg and charge

2 \times 10^{-6}

C moves with

\vec v = (4\hat i + 3\hat k)

m/s in a field

\vec B = \pi\hat k

T. Find its period, the radius of its helix and its pitch.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2026 · 8 Apr 2026 Shift 2 · Q23Moderate

Example 3 · Moving Charges and Magnetism · Circular and Helical Paths of Charges

A 5 mg particle carrying a charge of

5\pi \times 10^{- 6}C

is moving with velocity of

(3\widehat{i} + 2\widehat{k}) \times 10^{- 2}\text{ }m/s

in a region having magnetic field

\overrightarrow{B} = 0.1\widehat{k}Wb/m^{2}

. It moves a distance of

\alpha

meter along

\widehat{k}

when it completes 5 revolutions. The value of

\alpha

\_\_\_\_

Only the part of v along B makes the pitch

Pitch is v cos θ times the period. Using the full speed, or the part across B, gives a wrong distance along the field.

Two gains of energy per revolution in a cyclotron

The particle crosses the gap between the dees twice in each turn, gaining qV each time. Revolutions = K/(2qV); dividing by qV alone doubles the count.

The period does not depend on the speed

T = 2πm/qB has no v and no r in it. A faster charge goes round a larger circle in the same time, which is why the cyclotron frequency can stay fixed.

Drill 7 more on period of revolution, the helix and the cyclotron

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (3)

Radius of a charge's circular path and how it compares between particles
Radius of the circle
$r = \frac{mv}{qB} = \frac{p}{qB} = \frac{\sqrt{2mK}}{qB} = \frac{1}{B}\sqrt{\frac{2mV}{q}}$
Finding a radius, a mass or a field from r = mv/qB
Mass from a measured radius
$m = \frac{qBr}{v} = \frac{qB^{2}r^{2}}{2V}$
Period of revolution, the helix and the cyclotron
Period, pitch and cyclotron energy
$T = \frac{2\pi m}{qB} \qquad p = v\cos\theta\,T \qquad K_{\max} = \frac{q^{2}B^{2}R^{2}}{2m}$

Watch out for (9)

Equal energy and equal voltage are different conditions
→ Radius of a charge's circular path and how it compares between particles
Curvature is not radius
→ Radius of a charge's circular path and how it compares between particles
Radius goes as the square root of kinetic energy
→ Radius of a charge's circular path and how it compares between particles
Energy in eV must be turned into joules
→ Finding a radius, a mass or a field from r = mv/qB
Radius, not diameter
→ Finding a radius, a mass or a field from r = mv/qB
Singly ionised means charge e
→ Finding a radius, a mass or a field from r = mv/qB
Only the part of v along B makes the pitch
→ Period of revolution, the helix and the cyclotron
Two gains of energy per revolution in a cyclotron
→ Period of revolution, the helix and the cyclotron
The period does not depend on the speed
→ Period of revolution, the helix and the cyclotron

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