Physics Class 12 Part I…
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Magnetic Force

🎓 Class 12 Physics CBSE Theory Ch 4 – Moving Charges and Magnetism ⏱ ~14 min
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આ MCQ મોડ્યુલ આના પર આધારિત છે: Magnetic Force

આ મૂલ્યાંકન આના પર આધારિત હશે: Magnetic Force

મૂલ્યાંકન બનાવવામાં તેમની સામગ્રી સામેલ કરવા ચિત્રો, PDF અથવા Word દસ્તાવેજ અપલોડ કરો.

Magnetic Force

4.1 Introduction

You have already encountered a current-carrying wire producing an effect on a magnetic compass needle (Oersted's experiment, 1820). This was the historic moment when humanity discovered that electricity and magnetism are not separate phenomena - they are two sides of the same coin. In this chapter we study how moving charges produce magnetic fields and how magnetic fields, in turn, exert forces on moving charges.

We define the magnetic field through the force it exerts on a moving charge. The SI unit is the tesla (T), named after Nikola Tesla. One tesla is a very strong field; the Earth's field is only about \(0.5 \times 10^{-4}\) T (50 microtesla).

4.2 Magnetic Force

4.2.1 Sources and fields

Just as a static charge produces an electric field, a moving charge produces a magnetic field. The total electromagnetic force on a charge q moving with velocity \(\vec v\) in a region where both an electric field \(\vec E\) and a magnetic field \(\vec B\) exist is the Lorentz force.

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

4.2.2 Magnetic Force on a Moving Charge

The purely magnetic part is

\[\vec F_{mag} = q\,\vec v \times \vec B\]

Its magnitude is \(F = qvB\sin\theta\), where \(\theta\) is the angle between \(\vec v\) and \(\vec B\). Three crucial features of this force:

  1. It depends on q, \(\vec v\) and \(\vec B\). The force on a negative charge is opposite to that on a positive charge moving the same way.
  2. It includes a vector cross-product. The force is always perpendicular to both \(\vec v\) and \(\vec B\).
  3. If \(\vec v \parallel \vec B\) (or \(\vec v = 0\)), the magnetic force is zero. Maximum force occurs when \(\theta = 90^\circ\).
Key idea: Because \(\vec F\) is perpendicular to \(\vec v\), the magnetic force does no work on the charge. The kinetic energy (and hence the speed) of the charge stays constant - only its direction changes.
PALM B (fingers) v (thumb) F = qv x B (out of palm) Vector picture v B F (q>0)
Fig 4.1 Right-hand rule for the magnetic force on a positive charge: thumb points along v, fingers curl from v toward B; F emerges from the palm.
SI unit of B - the tesla: A field of one tesla exerts a force of one newton on a one-coulomb charge moving at one metre per second perpendicular to the field. Smaller field: 1 gauss = \(10^{-4}\) T.

Worked Example 4.1 - Force on an electron

Example 4.1 L3 Apply

An electron (\(q=-1.6\times10^{-19}\) C) moves east with speed \(v = 3\times10^7\) m s\(^{-1}\) through a magnetic field of \(B = 0.5\) T directed vertically upward. Find the magnetic force.

\(\vec v\) and \(\vec B\) are mutually perpendicular, so \(\sin\theta = 1\).

\(F = |q|vB = (1.6\times10^{-19})(3\times10^7)(0.5) = 2.4\times10^{-12}\) N.

By the right-hand rule, \(\vec v \times \vec B\) points south. Because the electron is negative, the force on it is to the north.

4.3 Motion in a Magnetic Field

If a charged particle enters a uniform magnetic field with its velocity perpendicular to \(\vec B\), the magnetic force always acts perpendicular to \(\vec v\) and so behaves like a centripetal force. The particle moves in a circle.

\(\dfrac{mv^2}{r} = qvB \quad\Longrightarrow\quad r = \dfrac{mv}{qB}\)

Period of one revolution:

\(T = \dfrac{2\pi r}{v} = \dfrac{2\pi m}{qB}\), and angular frequency \(\omega = \dfrac{qB}{m}\) (cyclotron frequency).

Note: T (and hence \(\omega\)) is independent of v and r - all particles of a given charge-to-mass ratio go round in the same time, no matter how fast.

If the velocity has a component along \(\vec B\) (call it \(v_\parallel\)), that component is unaffected by the magnetic force, so the particle traces a helix. Pitch (distance per turn) = \(v_\parallel T\).

B +q pitch p = v|| T
Fig 4.2 A charged particle with a velocity component along B traces a helical path.

Worked Example 4.2 - Radius and period

Example 4.2 L3 Apply

A proton (\(m=1.67\times10^{-27}\) kg, \(q=1.6\times10^{-19}\) C) moves perpendicular to a 0.20 T field with speed \(v = 4\times10^6\) m s\(^{-1}\). Find (a) the radius and (b) the period.

(a) \(r = \dfrac{mv}{qB} = \dfrac{(1.67\times10^{-27})(4\times10^6)}{(1.6\times10^{-19})(0.20)} = 0.209\) m \(\approx\) 21 cm.

(b) \(T = \dfrac{2\pi m}{qB} = \dfrac{2\pi(1.67\times10^{-27})}{(1.6\times10^{-19})(0.20)} = 3.28\times10^{-7}\) s.

4.4 Motion in Combined Electric and Magnetic Fields

4.4.1 Velocity Selector

If we apply mutually perpendicular E and B fields and send a charged beam at right angles to both, the electric force \(qE\) and the magnetic force \(qvB\) point in opposite directions. They cancel only when

\(qE = qvB \quad\Longrightarrow\quad v = \dfrac{E}{B}\)

So only particles with this special speed pass straight through - the device acts as a velocity selector. This was the principle of J. J. Thomson's e/m experiment.

+ + + + + + + - - - - - - - E B (out) v = E/B
Fig 4.3 Velocity selector. Only charges with v = E/B emerge undeflected.

4.4.2 Cyclotron

The cyclotron uses the speed-independence of the period T to repeatedly accelerate a charged particle to high energy.

Two semicircular metal boxes (called dees) sit in a uniform B field. An alternating voltage of frequency \(\nu_c = qB/(2\pi m)\) is applied across the gap. Each time the charge crosses the gap, the field has reversed direction so the particle is accelerated. Inside a dee, B alone bends it through a semicircle; the radius grows as v grows.

D1 D2 xx xx B into page ~ AC high-energy beam
Fig 4.4 Cyclotron: two D-shaped electrodes ("dees") in a perpendicular B field; AC across the gap repeatedly accelerates the particle on each crossing.

At maximum radius R (just before exit) the kinetic energy is

\(K = \dfrac{1}{2}mv^2 = \dfrac{q^2B^2R^2}{2m}\)

Limitation: at relativistic speeds the mass m grows with v, so T is no longer constant - the simple cyclotron fails. Modified machines (synchrocyclotron, synchrotron) overcome this.

Interactive: Cyclotron Radius Calculator L4 Analyse

Vary the magnetic field, particle mass and speed; see how the orbit radius and period change.

xxxxxx xxxxxx B (into page) r
r = 8.4 cm, T = 1.31e-7 s, K = 8.35e-14 J
Activity 4.1 - Right-Hand Rule with a Pencil L3 Apply

Hold a pencil horizontally in your right hand with the eraser pointing east (this is +v). Point your fingers north (this is +B). Your palm now faces upward.

Predict: which direction does the magnetic force point on a positive charge moving east through a northward field?

Up (out of palm). For a negative charge it would be down.

Now flip the velocity to west: the palm faces down so the force on +q points down. The cross-product reverses sign with either v or q.

Worked Example 4.3 - Cyclotron parameters

Example 4.3 L4 Analyse

A cyclotron with dee radius R = 0.60 m and B = 1.5 T accelerates protons. Find (a) the cyclotron frequency and (b) the maximum kinetic energy in MeV.

(a) \(\nu_c = \dfrac{qB}{2\pi m} = \dfrac{(1.6\times10^{-19})(1.5)}{2\pi(1.67\times10^{-27})} = 2.29\times10^7\) Hz \(\approx\) 22.9 MHz.

(b) \(K_{max} = \dfrac{q^2B^2R^2}{2m} = \dfrac{(1.6\times10^{-19})^2(1.5)^2(0.60)^2}{2(1.67\times10^{-27})} = 6.21\times10^{-12}\) J.

Convert: \(K = 6.21\times10^{-12} / 1.6\times10^{-13} \approx\) 38.8 MeV.

Competency-Based Questions L3-L5

A research team studies cosmic-ray protons entering Earth's magnetosphere (B \(\approx 5\times10^{-5}\) T). Some particles spiral down field lines and produce auroras; others bend so sharply that they cannot reach the surface near the equator.

Q1. Which expression gives the radius of a charged particle's circular path in a uniform magnetic field?

  • (a) r = qB/mv
  • (b) r = mv/qB
  • (c) r = qvB/m
  • (d) r = m/(qvB)
(b). From mv²/r = qvB.

Q2. (Short answer) Explain why the magnetic force never changes a charged particle's speed.

F = qv x B is always perpendicular to v, so F.v = 0 and the work done dW = F.v dt = 0. Kinetic energy and hence speed remain constant.

Q3. (Fill in the blank) The cyclotron frequency \(\omega = ...\) (in terms of q, B, m).

\(\omega = qB/m\). Independent of v and r.

Q4. (True or False) A velocity selector with E = 4 x 104 V/m and B = 0.10 T transmits charges of speed 4 x 105 m/s.

True. v = E/B = 4 x 104/0.10 = 4 x 105 m/s.

Q5. (HOT) Equator-trapping: explain why low-energy cosmic rays cannot reach the equator but easily reach the poles.

Near the equator B is roughly horizontal, perpendicular to incoming radial protons - they bend in tight circles and turn back. Near the poles B is nearly vertical (parallel to v), so the magnetic force is small and particles spiral down, ionising the atmosphere and producing auroras.

Assertion-Reason Questions L4 Analyse

(a) Both A and R true, R correctly explains A. (b) Both true, R does not explain A. (c) A true, R false. (d) A false, R true.

A: A magnetic field cannot change the speed of a charged particle.

R: The magnetic force is always perpendicular to the velocity.

(a). Perpendicular force does zero work; speed (KE) stays constant.

A: A proton and an electron entering the same field with the same speed travel in circles of equal radius.

R: Radius depends only on B and v.

(d). r = mv/qB depends on m too. The proton (1836 times heavier) makes a circle 1836 times larger.

A: A simple cyclotron cannot accelerate electrons to very high energies.

R: Electron mass varies appreciably (relativistic effect) at modest energies, breaking the resonance.

(a). Electrons reach relativistic speeds quickly; m increases, T grows and the AC frequency no longer matches.

Frequently Asked Questions - Magnetic Force

What is the main concept covered in Magnetic Force?
In NCERT Class 12 Physics Chapter 4 (Moving Charges and Magnetism), "Magnetic Force" covers the core principles and equations students need for board exam success. The MyAiSchool lesson explains the topic with definitions, derivations, worked examples, and interactive simulations. Key formulas and dimensional analysis are included to build conceptual depth and problem-solving skills aligned with the CBSE 2025-26 syllabus.
How is Magnetic Force useful in real-life applications?
Real-life applications of "Magnetic Force" from NCERT Class 12 Physics Chapter 4 include electronics, communication systems, medical imaging, solar energy, semiconductor devices, and modern technology. The MyAiSchool lesson links every concept to a tangible example so students see physics as a problem-solving framework for the physical world, not as abstract formulas.
What are the key formulas in Magnetic Force?
Key formulas in "Magnetic Force" (NCERT Class 12 Physics Chapter 4 Moving Charges and Magnetism) are derived step-by-step in the MyAiSchool lesson. Students should memorize the final formula AND understand its derivation for full board marks. Each formula is listed with its dimensional formula, SI unit, applicability range, and common pitfalls. The Summary section at the end of each part includes a quick-reference formula card.
How does this part connect to other parts of Chapter 4?
NCERT Class 12 Physics Chapter 4 (Moving Charges and Magnetism) is structured so each part builds on the previous one. "Magnetic Force" connects directly to neighbouring parts via shared definitions, units, and methodology. The MyAiSchool lesson cross-references related concepts with internal links so students can navigate the whole chapter as one connected story rather than disconnected fragments.
What types of CBSE board questions come from Magnetic Force?
CBSE board questions from "Magnetic Force" typically include: (1) 1-mark MCQs on definitions and formulas, (2) 2-mark short-answer derivations or applications, (3) 3-mark numerical problems with units, (4) 5-mark long-answer derivations followed by application. The MyAiSchool lesson tags each Competency-Based Question (CBQ) with Bloom level (L1-L6) so students know how to study for each weight.
How can students use the interactive simulation effectively?
The interactive simulation in the "Magnetic Force" lesson allows students to adjust input parameters (sliders or selectors) and see physical quantities update in real time. To use it effectively: (1) try extreme values to understand limiting cases, (2) compare with the analytical formula, (3) check unit consistency, (4) test special configurations from worked examples. The simulation reinforces conceptual intuition that pure formula manipulation cannot.
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Physics Class 12 Part I – NCERT (2025-26)
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