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JEE Mains Physics · Moving Charges and Magnetism

Lorentz Force and Crossed Fields

A charge moving through fields feels F = q(E + v × B); the magnetic part is always perpendicular to the velocity, so it bends the path but never changes the speed.

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Why this matters

Twenty-one PYQs, fifteen of them multiple choice, and four from 2026. Twelve are about the magnetic force alone: five work out q v × B or use the fact that the acceleration is perpendicular to B, five are statements about its direction and the fact that it does no work, and two put a charge near a current-carrying wire. Nine add an electric field: three are velocity selectors, three send a charge along a solenoid's axis, two ask when a charge can keep a constant velocity, and one finds the velocity from the total force.

Concept 1 of 2: Magnetic force on a moving charge, F = q v × B

A magnetic field pushes only on a moving charge, and it pushes sideways: at right angles both to the velocity and to the field. Because the push is always across the motion, it can turn the charge but never speed it up or slow it down. A charge moving along the field lines feels nothing at all.

Definition

  • F⃗=q(v⃗×B⃗)\vec F = q(\vec v \times \vec B), size qvBsin⁡θqvB\sin\theta, with θ\theta the angle between v⃗\vec v and B⃗\vec B.
  • Direction: v⃗×B⃗\vec v \times \vec B for a positive charge; reversed for an electron, F⃗=−e(v⃗×B⃗)\vec F = -e(\vec v \times \vec B).
  • Work it out by components: v⃗×B⃗=(vyBz−vzBy)i^−(vxBz−vzBx)j^+(vxBy−vyBx)k^\vec v \times \vec B = (v_yB_z - v_zB_y)\hat i - (v_xB_z - v_zB_x)\hat j + (v_xB_y - v_yB_x)\hat k.
  • F⃗⊥v⃗\vec F \perp \vec v and F⃗⊥B⃗\vec F \perp \vec B, so a⃗⋅B⃗=0\vec a \cdot \vec B = 0 and a⃗⋅v⃗=0\vec a \cdot \vec v = 0. No work is done: speed and kinetic energy stay constant.
  • v⃗∥B⃗\vec v \parallel \vec B: no magnetic force.
  • An electric force qE can change the speed; a magnetic force cannot.
  • A positive charge moving parallel to a long current, in the same direction, is pulled towards the wire; moving against the current, it is pushed away.

Magnetic force

F⃗=q(v⃗×B⃗),F=qvBsin⁡θ\vec F = q\left(\vec v \times \vec B\right), \qquad F = qvB\sin\theta

Worked example

A charge of 2 μC moves with v⃗=(2i^+3j^−k^)\vec v = (2\hat i + 3\hat j - \hat k) m/s in a field B⃗=(i^−j^+2k^)\vec B = (\hat i - \hat j + 2\hat k) T. Find the force on it and its size.
Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2026 · 2 Apr 2026 Shift 1 · Q23Moderate

Example 1 · Moving Charges and Magnetism · Lorentz Force and Crossed Fields

1μC1\mu C charge moving with velocity v→=(i^−2j^+3k^)m/s\overrightarrow{v} = (\widehat{i} - 2\widehat{j} + 3\widehat{k})m/s in the region of magnetic field B→=(2i^+3j^−5k^)\overrightarrow{B} = (2\widehat{i} + 3\widehat{j} - 5\widehat{k}) T. The magnitude of force acting on it is α×10−6 N\sqrt{\alpha} \times 10^{- 6}\text{ }N. The value of α\alpha is ____\_\_\_\_ .

v × B is not B × v

The cross product changes sign when the order is swapped. F = q v × B; writing B × v gives the force in exactly the opposite direction.

An electron's force is reversed

For an electron q = −e, so the force is along −(v × B). Finding the direction of v × B and stopping there gives the wrong answer for every electron question.

The magnetic force never changes speed

Because F is always perpendicular to v, it does no work. Speed and kinetic energy stay the same; only the direction of motion changes.
Drill 11 more on magnetic force on a moving charge, f = q v × b

Concept 2 of 2: Electric and magnetic forces together: the velocity selector

With an electric field as well, the total force is the Lorentz force, q(E + v × B). If E and B are perpendicular to each other and to the velocity, the two forces can point in opposite directions. At one particular speed, v = E/B, they cancel, and the charge goes straight through. Faster or slower charges are bent aside, which is why this arrangement is called a velocity selector.

Definition

  • Lorentz force: F⃗=q(E⃗+v⃗×B⃗)\vec F = q\left(\vec E + \vec v \times \vec B\right).
  • Velocity selector (E, B and v mutually perpendicular): undeflected when qE=qvBqE = qvB, that is v=E/Bv = E/B, whatever the charge or mass.
  • A charge given kinetic energy K has v=2K/mv = \sqrt{2K/m}; accelerated from rest through V, v=2qV/mv = \sqrt{2qV/m}.
  • Switch E off and the charge circles with r=mv/qBr = mv/qB (the next page); with B=E/vB = E/v this is r=mv2qEr = \dfrac{mv^{2}}{qE}.
  • A charge keeps a constant velocity if E = 0 and B = 0; if E = 0 and v is parallel to B; or if E and B are both present and their forces balance. It cannot with E ≠ 0 and B = 0.
  • Along the axis of a solenoid v is parallel to B, so there is no magnetic force; a charge there moves as it would with no field.

Lorentz force and the selected speed

F⃗=q(E⃗+v⃗×B⃗)v=EB\vec F = q\left(\vec E + \vec v \times \vec B\right) \qquad v = \frac{E}{B}

Worked example

Protons are accelerated from rest through 1250 V and then enter a region with a magnetic field of 0.06 T perpendicular to their path. What electric field, perpendicular to both, lets them pass undeflected? (proton mass 1.6×10−271.6 \times 10^{-27} kg, charge 1.6×10−191.6 \times 10^{-19} C)
Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2025 · 22 Jan 2025 · Q96Moderate

Example 2 · Moving Charges and Magnetism · Lorentz Force and Crossed Fields

A proton is moving undeflected in a region of crossed electric and magnetic fields at a constant speed of 2×105 ms−12 \times10^{5}{\text{ }ms}^{- 1}. When the electric field is switched off, the proton moves along a circular path of radius 2 cm . The magnitude of electric field is x×104 N/Cx \times10^{4}\text{ }N/C. the value of x is ______\_\_\_\_\_\_ Take the mass of the proton =1.6×10−27 kg= 1.6 \times10^{- 27}\text{ }kg.

The selector picks a speed, not a charge or a mass

v = E/B has no q and no m in it. A proton, an electron and an alpha particle at the same speed all pass straight through the same crossed fields.

Find v from the kinetic energy first

When a question gives an energy in eV, convert it to joules and use v = √(2K/m) before putting v into E = vB. Using the energy directly as a speed gives nonsense.

Constant velocity is impossible with only an electric field

An electric field always pushes a charge, whatever its motion. A region with E ≠ 0 and B = 0 cannot let a charge move at constant velocity; a region with B alone can, if the charge moves along B.
Drill 8 more on electric and magnetic forces together: the velocity selector

Summary — formulas & gotchas at a glance

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Formulas (2)

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