MHT-CET Physics · Magnetic Fields Due to Electric Current

Force on a Moving Charge, and Circular Motion in a Field

A charge q moving with velocity v through a magnetic field B feels F = q(v × B), perpendicular to both, so it does no work and only bends the path; moving across a uniform field the charge goes round a circle of radius mv/(qB), and adding an electric field gives the full Lorentz force qE + q(v × B).

All Magnetic Fields Due to Electric Current notes MHT-CET Physics strategy

Why this matters

12 PYQs, none HARD. Five are the force itself — its size from a cross product, the work it does (zero), a charge moving along the field lines, the full Lorentz force, what a cyclotron accelerates; seven are circular motion — the radius when the speed, field or kinetic energy changes, the radius after acceleration through a voltage, the mass from a semicircle, and an electron and a proton of equal momentum. Two cards.

Concept 1 of 2: The Magnetic Force on a Moving Charge

F = q(v × B): the force is perpendicular to the velocity, so it can turn the charge but never speed it up or slow it down — the work it does is always zero. Its size is qvB sin θ, so a charge moving along the field lines feels nothing. Work the cross product component by component when v and B are given as vectors. With an electric field too, the total is the Lorentz force qE + q(v × B). A cyclotron uses the magnetic force to bend and an alternating electric field to accelerate, so it works for any charged particle, positive or negative — but not for neutrons.

Definition

$\vec F = q(\vec v \times \vec B)$ , $|F| = qvB\sin\theta$ ; zero work (F ⟂ v); zero along the field lines.
$\vec v = a\hat i$ , $\vec B = b\hat j + c\hat k$ : $\vec F = qa(b\hat k - c\hat j)$ , $|F| = qa\sqrt{b^2 + c^2}$ .
Lorentz force: $\vec F = q\vec E + q(\vec v \times \vec B)$ .
Cyclotron: accelerates positive and negative charges, not neutrons.

Lorentz force

\vec F = q\vec E + q(\vec v \times \vec B)

Worked example

A proton moves with

v = (2\hat{i} + 3\hat{j}) \times 10^5

m/s in

B = 0.5\hat{k}

T. Force on it (e = 1.6 × 10⁻¹⁹ C)?

Practice this conceptself-check · 3 quick reps

The same idea in a real exam question:

MHT-CET · 2025 · 22 April Shift II · Q28Moderate

Example 1 · Magnetic Fields Due to Electric Current · Force on Moving Charge in Magnetic Field

A particle of charge

q

moves with a velocity

\overrightarrow{V}= a\widehat{i}

in a magnetic field

\overrightarrow{B}= b\widehat{j}+ c\widehat{k}

, where ' a ', ' b ' and ' c ' are constants. The magnitude of force experienced by particle is

Thinking a magnetic field can change a charge's speed

The magnetic force is perpendicular to the velocity at every instant, so it does no work: kinetic energy and speed stay fixed, only the direction turns.

Drill 4 more on the magnetic force on a moving charge

Concept 2 of 2: Circular Motion in a Magnetic Field

Moving straight across a uniform field, the magnetic force supplies the centripetal force: qvB = mv²/r, so r = mv/(qB) = p/(qB) = √(2mK)/(qB). Halve the speed and double the field and the radius falls to a quarter. The radius goes as the square root of the kinetic energy, so three times the energy gives √3 times the radius. A charge accelerated through V has K = qV, so r = (1/B)√(2mV/q). An electron and a proton with the same momentum and charge size move on circles of the same radius, turning opposite ways.

Definition

$r = \dfrac{mv}{qB} = \dfrac{p}{qB} = \dfrac{\sqrt{2mK}}{qB}$ .
v ÷ 2, B × 2 ⇒ r ÷ 4. K × 2 ⇒ $\sqrt{2}r$ ; K × 3 ⇒ $\sqrt{3}r$ .
Accelerated through V: $r = \dfrac{1}{B}\sqrt{\dfrac{2mV}{q}}$ ⇒ $m = \dfrac{qB^2r^2}{2V}$ .
Equal momenta, equal |q|: same radius, opposite senses.

Radius

r = \frac{mv}{qB} = \frac{\sqrt{2mK}}{qB}

Worked example

An alpha particle and a proton enter the same field with the same kinetic energy. Ratio of radii (alpha : proton)?

Practice this conceptself-check · 3 quick reps

The same idea in a real exam question:

MHT-CET · 2023 · 4th May Shift 1 · Q17Moderate

Example 2 · Magnetic Fields Due to Electric Current · Force on Moving Charge in Magnetic Field

A charged particle of charge

q

is accelerated by a potential difference

V

enters a region of uniform magnetic field

B

at right angles to the direction of field. The charged particle completes semicircle of radius

r

inside magnetic field. The mass of the charged particle is

Scaling the radius with the energy

r ∝ √K, not K. Doubling the kinetic energy makes the circle √2 times wider, not twice.

Drill 6 more on circular motion in a magnetic field

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 (2)

The Magnetic Force on a Moving Charge
Lorentz force
$\vec F = q\vec E + q(\vec v \times \vec B)$
Circular Motion in a Magnetic Field
Radius
$r = \frac{mv}{qB} = \frac{\sqrt{2mK}}{qB}$

Watch out for (2)

Thinking a magnetic field can change a charge's speed
→ The Magnetic Force on a Moving Charge
Scaling the radius with the energy
→ Circular Motion in a Magnetic Field

Test yourself on Magnetic Fields Due to Electric Current

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