Class 12 Physics · NCERT chapter 4 · 8 topics · free to read
Magnetic Force on Moving Charge
A moving charge in a magnetic field experiences a force perpendicular to both its velocity and the field. This Lorentz force is the engine behind everything from cyclotrons to MRI machines.
F = q( v B )
Lorentz Force
The magnetic force on a charge q moving with velocity v in a field B is given by their cross product. The force is always perpendicular to both v and B.
When a charge q moves with velocity v through a magnetic field B , it experiences a force called the Lorentz magnetic force. This force is the cross product of v and B , scaled by the charge.
F = q( v B ) (NCERT Eq. 4.2)
The magnitude of this force is F = qvB , where is the angle between v and B . The direction is given by the right-hand rule for cross products.
F = qvB
SI unit of magnetic field B → Tesla (T) = kg A^ -1 s^ -2
1 Tesla = 10^4 Gauss (Earth's field 0.5 Gauss = 5 10^ -5 T)
Force is ALWAYS perpendicular to velocity → does zero work on the charge
Since W = 0 , the magnetic force changes direction but NEVER changes speed or kinetic energy
For negative charges (electrons), the force direction reverses compared to the right-hand rule
Special Cases
The angle between v and B determines everything. Three critical cases: parallel, perpendicular, and at an arbitrary angle.
The magnitude F = qvB depends entirely on the angle between v and B . Three cases define all possible scenarios.
v B ( = 0° or 180° )
F = 0 → charge moves in a straight line
v B ( = 90° )
F = qvB (maximum) → charge moves in a circle
v at angle to B
F = qvB → charge moves in a helix
Quick recall
A proton enters a magnetic field at 30° to the field direction. What fraction of the maximum force does it experience?
F = qvB 30° = 1 2 qvB . It experiences half the maximum force and follows a helical path.
Combined Electric & Magnetic Force
When both E and B fields are present, the total force is the sum of electric and magnetic contributions.
In most real situations, both electric and magnetic fields coexist. The total Lorentz force on a charge combines both contributions.
F = q E + q( v B ) = q( E + v B ) (NCERT Eq. 4.3)
Acts on stationary AND moving charges
Acts ONLY on moving charges
Along or opposite to E
Perpendicular to both v and B
Does work on the charge
Does NO work on the charge
Can change speed
Can only change direction
Independent of velocity
Depends on velocity
Electric force exists even for stationary charges ( v = 0 ); magnetic force vanishes
A charge at rest in a magnetic field feels NO force
In a velocity selector: qE = qvB gives v = E/B — independent of charge and mass
Motion in Magnetic Field
A charged particle perpendicular to a uniform magnetic field traces a perfect circle. At an angle, it traces a helix. The radius, time period, and frequency of this motion are fundamental to particle physics.
r = mv qB
Circular Motion
The magnetic force acts as a centripetal force, bending the charge into a circular orbit. The radius depends on momentum and field strength.
When v B , the magnetic force F = qvB is always perpendicular to v . This means it acts as a centripetal force, bending the particle into a circle without changing its speed.
Setting the magnetic force equal to the centripetal force requirement gives us the radius of the circular orbit.
qvB = mv^2 r r = mv qB = p qB (NCERT Eq. 4.5)
r = mv/qB — radius increases with momentum, decreases with charge and field strength
r = p/qB — directly proportional to linear momentum p
For same v and B : heavier particle → larger radius, more charged particle → smaller radius
Proton and electron at same speed: r_p/r_e = m_p/m_e 1836 (proton makes a much bigger circle)
Kinetic energy form: r = 1 B 2mK q^2 where K is kinetic energy
Quick recall
If the speed of a proton in a magnetic field is doubled, what happens to the radius of its circular path?
Since r = mv/qB , doubling v doubles r . The circle becomes twice as large. But the time period stays the same!
Time Period & Cyclotron Frequency
The time to complete one circle and the frequency of revolution do NOT depend on the speed of the particle — a remarkable result that makes the cyclotron possible.
The time period T is the circumference 2 r divided by speed v . Substituting r = mv/qB gives a stunning result — v cancels out completely.
T = 2 r v = 2 m qB (NCERT Eq. 4.6)
f = 1 T = qB 2 m ; = qB m
T depends only on m , q , and B — NOT on v or r
Cyclotron frequency _c = qB/m is also called gyrofrequency
Heavier particles orbit slower (larger T ); more charged particles orbit faster (smaller T )
This independence of speed is what makes the cyclotron work — the alternating voltage frequency stays fixed
Helical Motion
When v is at an angle to B, the parallel component gives straight-line motion while the perpendicular component gives circular motion. The combination is a helix (corkscrew).
When v makes angle with B , decompose into: v_ = v (along B ) and v_ = v (perpendicular to B ). The parallel component is unaffected; the perpendicular component causes circular motion.
r = mv qB ; Pitch p = v T = 2 m v qB
At = 90° : pure circular motion (pitch = 0)
At = 0° : straight line (no circular component, no force)
At any other angle: helical motion with pitch p = 2 mv / (qB)
The helix radius depends on v ; the pitch depends on v
Quick recall
A proton enters a magnetic field at 60° to the field. What is the ratio of radius to pitch of the helix?
The cyclotron uses the fact that charged particles orbit at a fixed frequency in a magnetic field, regardless of speed. By syncing an alternating voltage to this frequency, particles are accelerated to enormous energies in a compact device.
f_c = qB 2 m
Cyclotron Principle
The cyclotron exploits the independence of orbital frequency from speed. An oscillating electric field gives the particle a 'kick' every half-revolution, and the particle spirals outward at ever-increasing speed.
The cyclotron was invented by E.O. Lawrence in 1934. It uses two hollow D-shaped electrodes (dees) placed in a uniform magnetic field. A high-frequency oscillating voltage is applied across the gap between the dees.
The key principle: since T = 2 m/(qB) is independent of speed, the oscillator frequency can be set to match the particle's orbital frequency. Every time the particle crosses the gap, it gets accelerated by the electric field.
The cyclotron fails for electrons (too light) and for very high energies (relativistic mass increase). Neutral particles cannot be accelerated at all.
The cyclotron has three important limitations that determine what it can and cannot do.
Cannot accelerate electrons — electrons are so light that they reach relativistic speeds very quickly, and their mass increases, disrupting the resonance condition
Cannot accelerate neutral particles — both the magnetic force ( qv B ) and the electric force ( qE ) require charge; neutral particles feel neither
Relativistic limit — at speeds close to c , the mass increases as m = m_0/ 1-v^2/c^2 , so T increases and the particle falls out of sync with the oscillator
Biot-Savart Law
The Biot-Savart Law tells you the magnetic field produced by a tiny element of current-carrying wire. It is to magnetism what Coulomb's Law is to electrostatics — the fundamental building block for calculating B-fields.
d B = _0 4 I d l r r^2
Biot-Savart Law
Each tiny segment of a current-carrying wire produces a tiny magnetic field. The Biot-Savart Law gives the magnitude and direction of this field at any point in space.
A current element Id l at a point produces a magnetic field d B at a field point located at position vector r from the element.
d B = _0 4 I d l r r^2 (NCERT Eq. 4.11)
|dB| = _0 4 I dl r^2
Here is the angle between d l (direction of current) and r (direction from element to field point). _0 = 4 10^ -7 T·m/A is the permeability of free space.
dB I — field is proportional to current
dB dl — field is proportional to element length
dB — maximum when = 90° , zero when = 0° or 180°
dB 1/r^2 — inverse square law, just like Coulomb's Law
Direction of d B : perpendicular to the plane containing d l and r (use right-hand rule)
dB = _0 4 Idl r^2
E = 1 4 _0 q r^2
Source: current element Idl
Source: point charge q
Inverse square law: 1/r^2
Inverse square law: 1/r^2
Field direction: perpendicular to plane of dl and r
Field direction: along r
Principle of superposition holds
Principle of superposition holds
Magnetic Field Due to a Straight Wire
Applying Biot-Savart to a finite straight wire gives a formula involving the angles subtended at the field point. For an infinite wire, this simplifies beautifully.
For a finite straight wire carrying current I , the field at perpendicular distance d from the wire involves the angles _1 and _2 subtended by the two ends of the wire at the point.
B = _0 I 4 d ( _1 + _2)
For an infinitely long straight wire, both angles are 90° , so _1 + _2 = 2 .
B = _0 I 2 d (Infinite straight wire — NCERT Eq. 4.19)
For a semi-infinite wire (one end at the point): B = _0 I/(4 d) (half of infinite wire)
The field forms concentric circles around the wire — use the right-hand thumb rule
Right-hand thumb rule: thumb along current → fingers curl in the direction of B
The field decreases as 1/d (NOT 1/d^2 ) — this is because the wire is one-dimensional, not a point
Magnetic Field of a Circular Loop
A circular current loop produces a field that looks like a magnetic dipole. At the center, the formula is simple. On the axis, it involves the geometry of the loop.
For a circular loop of radius R carrying current I with N turns, the magnetic field at the center of the loop is:
B_ center = _0 N I 2R (NCERT Eq. 4.16)
On the axis of the loop at distance x from the center:
B_ axis = _0 N I R^2 2(R^2 + x^2)^ 3/2 (NCERT Eq. 4.15)
At center ( x = 0 ): B = _0 NI/(2R) — maximum value
Far away ( x R ): B _0 NIA/(2 x^3) where A = R^2 — falls as 1/x^3 (dipole behavior)
Direction at center: use right-hand curl rule — curl fingers along current, thumb gives B direction
Field lines of a current loop look exactly like those of a bar magnet — this is no coincidence
Quick recall
A circular coil of 100 turns, radius 10 cm, carries 2 A. What is the field at the center?
B = _0 NI/(2R) = 4 10^ -7 100 2/(2 0.1) = 4 10^ -4 T 1.26 10^ -3 T = 1.26 mT
Ampere's Circuital Law
Ampere's Law relates the line integral of the magnetic field around a closed loop to the total current enclosed. It is the magnetic equivalent of Gauss's Law — powerful when symmetry is present.
B d l = _0 I_ enc
Ampere's Circuital Law
The line integral of B around any closed loop (Amperian loop) equals μ₀ times the net current threading through that loop. Choosing the right loop simplifies hard problems to one-liners.
Ampere's Circuital Law states that the line integral of the magnetic field B around any closed path (called an Amperian loop) equals _0 times the total current enclosed by that path.
B d l = _0 I_ enc (NCERT Eq. 4.17)
I_ enc is the NET current passing through the area bounded by the Amperian loop
Current going one way is positive; opposite direction is negative (use right-hand rule)
The law is ALWAYS true, but only USEFUL when B has enough symmetry to be pulled out of the integral
Best for: infinite straight wires, solenoids, toroids — all with high symmetry
B d l = _0 I_ enc
E d A = Q_ enc / _0
Line integral around a closed loop
Surface integral over a closed surface
Source: current ( I )
Source: charge ( Q )
Useful when B is constant on the loop
Useful when E is constant on the surface
Applies to wires, solenoids, toroids
Applies to spheres, planes, cylinders
Solenoid
A solenoid is a tightly wound coil of wire. Inside, the field is remarkably uniform and strong. Outside, it is nearly zero. Ampere's Law gives the field in two lines.
An ideal solenoid (tightly wound, infinite length) has a perfectly uniform field inside and zero field outside. Apply Ampere's Law to a rectangular Amperian loop partly inside, partly outside.
B l = _0 (n l) I B = _0 n I (NCERT Eq. 4.20)
Here n = N/L is the number of turns per unit length. The field depends only on n and I — not on the position inside the solenoid or its cross-sectional area.
B = _0 n I — uniform everywhere inside an ideal solenoid
B = 0 outside an ideal solenoid (field lines are confined inside)
n = N/L — total turns divided by length
At the ends of a finite solenoid: B_ end = 1 2 _0 n I (exactly half)
The field direction inside: use the right-hand rule or the SNOW rule
Quick recall
A solenoid has 2000 turns over 50 cm and carries 3 A. What is the field inside?
n = 2000/0.5 = 4000 turns/m. B = _0 nI = 4 10^ -7 4000 3 = 4.8 10^ -3 T 0.015 T = 15 mT.
Toroid
A toroid is a solenoid curved into a doughnut shape. The field is entirely confined inside the ring — zero outside and zero in the central hole.
A toroid is a solenoid bent into a closed ring (doughnut shape). It has N total turns and mean radius R . Applying Ampere's Law with a circular Amperian loop of radius r centered on the toroid axis:
B(2 r) = _0 N I B = _0 N I 2 r (NCERT Eq. 4.22)
Inside the toroid (within the windings): B = _0 NI/(2 r)
In the central hole ( r < inner radius): B = 0 (no current enclosed)
Outside the toroid ( r > outer radius): B = 0 (currents cancel — equal current in and out)
For thin toroid: B _0 nI where n = N/(2 R) — same as solenoid formula
Straight cylinder
Bent into a ring (doughnut)
B = _0 nI (uniform inside)
B = _0 NI/(2 r) (varies slightly with r )
Field leaks from the ends
No leakage — field completely confined
B 0 outside (for finite solenoid)
B = 0 everywhere outside
Infinite solenoid: truly uniform
Thin toroid: approximately uniform
Force Between Current-Carrying Conductors
Two parallel current-carrying wires exert forces on each other through their magnetic fields. Parallel currents attract, anti-parallel currents repel. This interaction defines the SI unit of current — the Ampere.
F L = _0 I_1 I_2 2 d
Force Between Parallel Wires
Wire 1 creates a magnetic field at Wire 2's location. Wire 2, carrying current in this field, experiences a force. The direction depends on whether the currents are in the same or opposite directions.
Wire 1 carrying current I_1 creates a field B_1 = _0 I_1/(2 d) at the location of Wire 2 (distance d away). Wire 2, carrying current I_2 in this field, experiences a force per unit length.
F L = _0 I_1 I_2 2 d (NCERT Eq. 4.23)
Parallel currents (same direction) → force is attractive
Anti-parallel currents (opposite direction) → force is repulsive
Force per unit length I_1 I_2 / d — proportional to both currents, inversely proportional to distance
The force on Wire 1 due to Wire 2 equals the force on Wire 2 due to Wire 1 (Newton's third law)
This force is mutual — both wires feel the same magnitude of force
Quick recall
Two parallel wires 10 cm apart carry 5 A each in the same direction. What is the force per meter?
F/L = _0 I_1 I_2/(2 d) = 4 10^ -7 5 5/(2 0.1) = 5 10^ -5 N/m. The force is attractive.
Definition of Ampere
The Ampere is defined using the force between two parallel wires. This makes current a fundamental quantity measured through its magnetic effect.
The SI unit of current, the Ampere, was historically defined using the magnetic force between parallel wires.
F L = _0 1 1 2 1 = 4 10^ -7 2 = 2 10^ -7 N/m
1 Ampere → force of 2 10^ -7 N/m between wires 1 m apart
This definition makes current a base quantity (not derived from charge)
1 Coulomb = 1 Ampere 1 second (charge is DEFINED from current, not the other way around)
Since 2019, the Ampere is redefined using the elementary charge e = 1.602176634 10^ -19 C exactly — but NCERT uses the older definition
Force & Torque on Current Loop
A current-carrying conductor in a magnetic field experiences a force (F = Il×B). A current loop experiences a torque that tries to align its magnetic moment with the field — this is the principle behind electric motors.
= m B
Force on a Current-Carrying Conductor
A straight wire of length L carrying current I in a field B experiences a force. This is just the sum of Lorentz forces on all the moving charges inside the wire.
A straight conductor of length L carrying current I in a uniform magnetic field B experiences a net force. This is the macroscopic result of adding up the Lorentz forces on all the drifting electrons.
F = I( L B ) (NCERT Eq. 4.4)
F = BIL
L is a vector along the wire in the direction of current, and is the angle between L and B .
When L B ( = 90° ): F = BIL — maximum force
When L B ( = 0° ): F = 0 — no force
Direction: use Fleming's Left-Hand Rule (FBI: Force, B-field, Current)
This is the principle behind every electric motor — current in a magnetic field produces force
Magnetic Dipole Moment
A current loop behaves like a tiny bar magnet. Its magnetic dipole moment m = NIA captures the strength and direction of this 'magnet.'
A planar current loop with N turns, each of area A , carrying current I , has a magnetic dipole moment m .
m = NI A ; |m| = NIA (NCERT Eq. 4.28)
The direction of m is along the normal to the loop, given by the right-hand curl rule: curl fingers along the current direction, and the thumb points along m .
SI unit of m → A·m² (Ampere metre-squared)
m is perpendicular to the plane of the loop
For N turns: m = NIA — each turn contributes IA
A current loop's far-field looks exactly like a bar magnet — the loop IS a magnetic dipole
m is independent of shape — circular, square, or any planar loop with area A
Torque on a Current Loop
When a current loop is placed in a uniform magnetic field, opposite forces on opposite arms create a torque. This torque tries to align the magnetic moment with the field.
In a uniform magnetic field, the net force on a current loop is zero (forces on opposite sides cancel). But the forces create a torque (couple) that tends to rotate the loop.
= m B = NIA B (NCERT Eq. 4.29)
Here is the angle between the magnetic moment m (normal to the loop) and B .
Maximum torque ( = NIAB ) when m B ( = 90° ) — loop plane is parallel to B
Zero torque when m B ( = 0° ) — loop plane is perpendicular to B (stable equilibrium)
Zero torque also when m antiparallel to B ( = 180° ) — but this is UNSTABLE equilibrium
The torque tries to align m with B — just like a compass needle aligning with Earth's field
= p E
= m B
U = - p E
U = - m B
Stable at p E
Stable at m B
Unstable at p antiparallel E
Unstable at m antiparallel B
U = - m B = -mB (Potential Energy)
Quick recall
A 50-turn coil of area 20 cm² carrying 0.5 A is placed with its plane at 30° to a 0.4 T field. What is the torque?
The angle between m and B is 90° - 30° = 60° (since m is normal to the plane). = NIAB 60° = 50 0.5 20 10^ -4 0.4 ( 3 /2) = 0.0173 N·m.
Moving Coil Galvanometer
The galvanometer is a sensitive instrument that detects and measures small currents using the torque on a current loop in a magnetic field. By adding resistances, it becomes an ammeter or voltmeter.
I = k NAB
Principle & Construction
A coil suspended in a radial magnetic field experiences a torque proportional to the current. A spring provides a restoring torque. The deflection is directly proportional to current.
A rectangular coil of N turns and area A is suspended in a radial magnetic field B (created by a concentric soft-iron core). When current I flows, the coil experiences torque = NIAB and deflects until the restoring spring torque k balances it.
NIAB = k I = k NAB (NCERT Eq. 4.31)
Since I , the scale is linear — equal current increments produce equal deflection increments. This linearity comes from the radial magnetic field, which ensures = 1 for all positions.
Radial magnetic field: B is always parallel to the plane of the coil → _ effective always = 90° → = NIAB always
Soft-iron core: makes the field radial AND increases field strength (high permeability)
Phosphor-bronze spring: provides restoring torque _r = k AND carries current to the coil
Deflection = NAB I/k → linear scale
Sensitivity
Current sensitivity is deflection per unit current. Voltage sensitivity is deflection per unit voltage. Increasing one may not increase the other.
Current sensitivity S_I is the deflection produced per unit current. Voltage sensitivity S_V is the deflection per unit voltage applied.
S_I = I = NAB k ; S_V = V = NAB kR
To increase S_I : increase N , A , B or decrease k (use softer spring)
Increasing N increases S_I but also increases resistance R → S_V may not increase
S_V = S_I / R — voltage sensitivity depends on resistance too
A galvanometer with high current sensitivity may have low voltage sensitivity if R is large
Conversion to Ammeter
A galvanometer becomes an ammeter by connecting a low resistance (shunt) in parallel. The shunt diverts most of the current, protecting the sensitive coil.
A galvanometer has high resistance G and can only handle small currents I_g (full-scale deflection). To measure large currents, connect a low resistance S (called a shunt) in parallel.
S = I_g G I - I_g (NCERT Eq. 4.33)
The shunt carries (I - I_g) while the galvanometer carries I_g . Since they are in parallel, I_g G = (I - I_g)S .
Shunt is in parallel → ammeter resistance R_A = GS/(G+S) S (very low)
An ideal ammeter has zero resistance — shunt brings it close to zero
Ammeter is connected in series with the circuit element
Low resistance ensures the ammeter doesn't significantly change the current it's measuring
Quick recall
A galvanometer with G = 100 and I_g = 1 mA is to be converted to a 10 A ammeter. Find the shunt.
S = I_g G/(I - I_g) = 0.001 100/(10 - 0.001) 0.01 . The shunt is extremely small — nearly all current bypasses the coil.
Conversion to Voltmeter
A galvanometer becomes a voltmeter by connecting a high resistance in series. This limits the current through the coil and allows it to measure large voltages.
To measure voltage, the galvanometer needs to be connected across (in parallel with) the circuit element. A high resistance R in series limits the current to I_g when the full voltage V is applied.
R = V I_g - G (NCERT Eq. 4.34)
High resistance is in series with galvanometer → total resistance R_V = G + R (very high)
An ideal voltmeter has infinite resistance — series resistance brings it close to infinity
Voltmeter is connected in parallel with the circuit element
High resistance ensures the voltmeter draws negligible current from the circuit
Low resistance shunt in parallel
High resistance in series
Connected in series with element
Connected in parallel with element
Ideal resistance: zero
Ideal resistance: infinite
S = I_g G/(I-I_g)
R = V/I_g - G
Measures current (A)
Measures potential difference (V)
Quick recall
A galvanometer with G = 50 and I_g = 2 mA is to be converted to a 0–10 V voltmeter. Find the series resistance.
R = V/I_g - G = 10/0.002 - 50 = 5000 - 50 = 4950 4.95 k .