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Class 12 Physics Notes

Moving Charges and Magnetism Class 12 Physics Notes

Complete, exam-ready notes on moving charges and magnetism: Lorentz force, motion of a charge in a magnetic field, Biot-Savart law, Ampere's circuital law, solenoid and toroid, force between currents, circular loop, galvanometer and the cyclotron — written for CBSE boards, JEE and NEET revision.

Class12SubjectPhysicsCoversCBSE · JEE · NEET

Written byDeep Narayan· Science & Mathematics EducatorReviewed byPushpanjali

What is the Lorentz force in one line?

The Lorentz force is the total force on a charged particle due to electric and magnetic fields: F = qE + q(v × B). The magnetic part acts perpendicular to both velocity and field, so it changes only the direction of motion, never the speed.

Lorentz Force and Motion in a Magnetic Field

Lorentz force

F⃗=qE⃗+q(v⃗×B⃗)\vec{F} = q\vec{E} + q(\vec{v}\times\vec{B})

The magnetic force is always perpendicular to velocity, doing zero work. It causes a charged particle to move in a circular path of radius r=mvqBr = \frac{mv}{qB} with angular frequency ω=qBm\omega = \frac{qB}{m}.

r=mvqB,ω=qBm,T=2πmqBr = \frac{mv}{qB},\quad \omega = \frac{qB}{m},\quad T = \frac{2\pi m}{qB}
Circular motion of a charge in a field

Helical path

When the velocity has a component parallel to the field, the charge moves in a helical path: circular motion from the perpendicular component plus constant drift along the field direction.

Biot-Savart Law

Biot-Savart law

dB=μ04πI dlsin⁡θr2dB = \frac{\mu_0}{4\pi}\frac{I\,dl\sin\theta}{r^2}

Gives the magnetic field produced by a small current element. For a long straight conductor, the field at distance a is B=μ0I2πaB = \frac{\mu_0 I}{2\pi a}.

  • Field at the centre of a circular loop:
  • B=μ0I2RB = \frac{\mu_0 I}{2R}
  • .
  • Field of a circular loop at a point on the axis:
  • B=μ0IR22(R2+x2)3/2B = \frac{\mu_0 I R^2}{2(R^2 + x^2)^{3/2}}
  • .
  • Right-hand grip rule gives the direction (curl fingers along current).
Bstraight=μ0I2πa,Bcentre=μ0I2RB_{\text{straight}} = \frac{\mu_0 I}{2\pi a},\quad B_{\text{centre}} = \frac{\mu_0 I}{2R}
Fields from a straight wire and a loop

Ampere's Circuital Law

Ampere's law

∮B⃗⋅dl⃗=μ0Ienclosed\oint \vec{B}\cdot d\vec{l} = \mu_0 I_{\text{enclosed}}

The line integral of the magnetic field around a closed loop equals μ₀ times the net current enclosed. It is used to find the field of symmetric current distributions.

  • Solenoid field inside:
  • B=μ0nIB = \mu_0 n I
  • (n = turns per unit length).
  • Toroid field:
  • B=μ0NI2πrB = \frac{\mu_0 NI}{2\pi r}
  • .
  • The field outside an ideal solenoid is zero.
Bsolenoid=μ0nI,Φ=BAcos⁡θB_{\text{solenoid}} = \mu_0 n I,\quad \Phi = BA\cos\theta
Solenoid field and magnetic flux

Force Between Two Parallel Currents

Force between parallel wires

F=μ0I1I2L2πdF = \frac{\mu_0 I_1 I_2 L}{2\pi d}

Parallel currents in the same direction attract; opposite directions repel. This defines the ampere: two parallel conductors 1 m apart carrying 1 A attract with a force of 2×10−7 N m−12\times10^{-7}\,\text{N m}^{-1}.

Moving Coil Galvanometer

Galvanometer deflection

θ=NBAkI\theta = \frac{NBA}{k}I

Deflection is proportional to current, with N turns of area A in a field B and spring constant k. It detects small currents and is the basis of ammeters and voltmeters.

  • To make an ammeter: connect a small shunt resistance in parallel.
  • To make a voltmeter: connect a large series resistance.
  • A radial magnetic field keeps the torque independent of the coil position, giving a linear scale.

Cyclotron

A cyclotron accelerates charged particles using a perpendicular magnetic field (which bends them in circles) and an alternating electric field across two dees. Since the angular frequency ω = qB/m is independent of speed, particles stay in resonance as they gain energy.

Relativistic limit

At very high speeds, the cyclotron fails because the mass of the particle increases relativistically, breaking the constant-frequency resonance. Synchrotrons overcome this.

Solved Examples

Example: A proton moves with v=2×106 m/sv = 2\times10^6\,\text{m/s} in a B=0.5 TB = 0.5\,\text{T} field perpendicular to its velocity. Find the radius (mp=1.67×10−27 kgm_p = 1.67\times10^{-27}\,\text{kg}).

Solution: r=mvqB=1.67×10−27×2×1061.6×10−19×0.5≈0.042 m=4.2 cmr = \frac{mv}{qB} = \frac{1.67\times10^{-27}\times2\times10^6}{1.6\times10^{-19}\times0.5} \approx 0.042\,\text{m} = 4.2\,\text{cm}.

Revision

Key formulas at a glance

Memorise these before attempting numericals — most exam questions hinge on one of them.

Lorentz force

F=q(E+v×B)F = q(E + v\times B)

Circular radius

r=mvqBr = \frac{mv}{qB}

Angular frequency

ω=qBm\omega = \frac{qB}{m}

Straight wire field

B=μ0I2πaB = \frac{\mu_0 I}{2\pi a}

Loop centre field

B=μ0I2RB = \frac{\mu_0 I}{2R}

Solenoid field

B=μ0nIB = \mu_0 n I

Force between wires

F=μ0I1I2L2πdF = \frac{\mu_0 I_1 I_2 L}{2\pi d}

Galvanometer

θ=NBAkI\theta = \frac{NBA}{k}I

Exam tips

How this chapter is asked

Where this topic appears in CBSE, JEE Main and NEET papers.

  • Magnetic force does no work — it changes direction, not speed.
  • Cyclotron frequency ω = qB/m is independent of velocity.
  • Parallel currents attract; anti-parallel repel.
  • Ampere's law: ∮B·dl = μ₀I_enc.
  • Biot-Savart gives dB; integrate for extended wires and loops.

FAQ

Common questions

Why does a magnetic field do no work on a moving charge?

The magnetic force F = q(v × B) is always perpendicular to the velocity, so its dot product with displacement is zero — it changes only direction, never speed or energy.

What is the difference between Biot-Savart law and Ampere's law?

Biot-Savart law computes the field from a current element by integration; Ampere's law uses symmetry (∮B·dl = μ₀I_enc) to find fields of highly symmetric distributions faster.

How is a galvanometer converted into an ammeter?

By connecting a small shunt resistance in parallel, which diverts most of the current, letting the galvanometer measure a small calibrated fraction of the total current.

Why are parallel currents attracting?

Each current produces a magnetic field that exerts a Lorentz force on the other. By the right-hand rule, same-direction currents produce an attractive force between them.

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Related notes

All Class 12 NotesClass 12 Current Electricity NotesClass 12 Electric Charges and Fields NotesPhysics Tutor in Mohali

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