NCERT Solutions for Class 12 Physics Chapter 4: Moving Charges and Magnetism
All 13 exercise questions of the latest NCERT textbook (2026–27 reprint) solved like a topper’s answer-sheet — magnetic field of coils, wires and solenoids, force on current-carrying conductors and between parallel currents, torque on a loop, the moving coil galvanometer and motion of a charge in a magnetic field — with diagrams linked to the exact NCERT pages, comparison tables, complete revision notes and exam tips.
Chapter at a glance — complete revision notes
4.1 Introduction
- Electricity and magnetism were known for more than 2000 years, but only in 1820 was their intimate relation discovered: during a lecture demonstration the Danish physicist Hans Christian Oersted (1777–1851) noticed that a current in a straight wire caused a noticeable deflection of a nearby magnetic compass needle. He found that the needle turns to align along the tangent to concentric circles round the wire [Fig. 4.1(c)], and that the deflection reverses when the current reverses. Moving charges (currents) produce magnetic fields.
- Soon after, Ampère, Biot, Savart and others developed the mathematical theory; the unified science is called electromagnetism. This chapter studies the magnetic force on moving charges and current-carrying conductors and the magnetic field produced by currents (Biot–Savart law, Ampere’s law), and ends with the moving coil galvanometer. Applications mentioned include the cyclotron.
4.2 Magnetic force
4.2.1 Sources and fields
Electric fields are produced by charges; magnetic fields by moving charges/currents (and by intrinsic magnetic moments of particles). Fields exist in space; a charge or current in the field experiences a force. Fig. 4.1: iron filings sprinkled around a long straight wire form concentric circles — the field is circular around the wire; its direction is given by the right-hand thumb rule (thumb along the current, curled fingers show the field direction).4.2.2 Magnetic field, Lorentz force
A charge q moving with velocity v in a magnetic field B (and electric field E) experiences the Lorentz force F = q (v × B) + q E; the magnetic part F = q v B sin θ is perpendicular to both v and B (direction by the right-hand/screw rule; reversed for a negative charge) and it is zero if v ∥ B or v = 0. Since F ⟂ v the magnetic force does no work: it changes the direction but not the speed or kinetic energy of the particle. Unit of B: tesla (T) = N s C⁻¹ m⁻¹ = N A⁻¹ m⁻¹ (after Nikola Tesla, 1856–1943); a smaller unit is the gauss (1 G = 10⁻⁴ T); Earth’s field ≈ 3.6 × 10⁻⁵ T. A magnetic field is defined by this force.4.2.3 Force on a current-carrying conductor
For a straight rod of length l with current I (n carriers per unit volume, drift velocity v_d): F = I l × B, magnitude F = I l B sin θ; direction given by the right-hand rule (Fig. 4.2). For a wire of arbitrary shape, F = I ∫ dl × B. Example 4.1: a straight wire of mass 200 g and length 1.5 m with 2 A is suspended mid-air by a horizontal field: mg = IlB ⇒ B = 0.65 T. Example 4.2: B along +y and velocity along +x: v × B is along +z, so the force on a proton is along +z and on an electron along −z.
4.3 Motion in a magnetic field
- Because the magnetic force does no work, a charge moving with v ⟂ B follows a circle (Fig. 4.5): q v B = m v²/r ⇒ r = m v/(q B) = p/(qB); larger momentum ⇒ larger circle. Angular frequency ω = qB/m, frequency ν = qB/(2π m) — the cyclotron frequency, independent of the speed and the radius (used in the cyclotron), period T = 2πm/qB.
- If v has a component along B (v∥), it is unaffected (no force along B), so the path is a helix (Fig. 4.6) with pitch p = v∥ T = 2π m v∥/(qB) and helix radius r = m v⊥/(qB). Example 4.3: electron (9 × 10⁻³¹ kg, 1.6 × 10⁻¹⁹ C) at 3 × 10⁷ m/s in 6 × 10⁻⁴ T: r = 28 cm, ν = 17 MHz, energy = 40.5 × 10⁻¹⁷ J ≈ 2.5 keV.
4.4 Magnetic field due to a current element — Biot–Savart law
- For an element dl carrying current I, the field at a point P at displacement r (angle θ between dl and r) (Fig. 4.7): dB = (μ₀/4π) · I dl sin θ / r², in vector form dB = (μ₀/4π) I (dl × r)/r³; direction perpendicular to the plane of dl and r (right-hand rule). μ₀ = 4π × 10⁻⁷ T m A⁻¹ is the permeability of free space.
- Comparison with Coulomb’s law: both are long-range, inverse-square; but the electric field due to a charge element is along r (radial) while the magnetic field is perpendicular to both dl and r (a vector product); the current element I dl cannot be isolated (a steady current flows in a closed loop), and there is an angle dependence (sin θ) — field is zero along the line of the element. Example 4.4: an element Δx = 1 cm carrying 10 A at the origin gives, at 0.5 m on the y-axis, B = (μ₀/4π) I Δx/r² = 10⁻⁷ × 10 × 10⁻²/0.25 = 4 × 10⁻⁸ T, perpendicular to the plane containing the element and the y-axis.
4.5 Magnetic field on the axis of a circular current loop
- For a loop of radius R carrying I, on its axis at distance x from the centre (Fig. 4.9): B = μ₀ I R² / [2 (x² + R²)^(3/2)]. At the centre (x = 0): B = μ₀ I/(2R); for N turns B = μ₀NI/2R. For x ≫ R: B ≈ μ₀ I R²/(2x³) = μ₀ m/(2π x³), where m = I π R² is the magnetic moment — like a dipole.
- Field lines (Fig. 4.10) of a current loop: closed curves; near the centre they are nearly straight, parallel to the axis. Right-hand rule for a loop: curl the fingers of the right hand along the current; the thumb gives the direction of B on the axis. Example 4.5: semicircular arc of radius 2.0 cm carrying 12 A (Fig. 4.11): straight segments contribute zero at the centre (dl ∥ r); the arc gives B = μ₀I/4R = 1.9 × 10⁻⁴ T (half that of a full loop) — direction depends on whether the wire bends one way or the other. Example 4.6: 100 turns, radius 10 cm, 1 A: B = μ₀NI/2R = 6.28 × 10⁻⁴ T.
4.6 Ampere’s circuital law
- Statement: the line integral of B around any closed loop equals μ₀ times the total current passing through the surface bounded by the loop: ∮ B · dl = μ₀ I_enclosed (sign of I by the right-hand rule). It is the magnetic analogue of Gauss’s law and, like Gauss’s law, is most useful for symmetric current distributions — it is not independent of the Biot–Savart law (it can be derived from it).
- Long straight wire: loop = circle of radius r: B × 2πr = μ₀ I ⇒ B = μ₀ I/(2π r) (tangential; field lines are concentric circles).
- Example 4.7 (wire of radius a with uniform current): outside (r > a): B = μ₀ I/2π r; inside (r < a): B = μ₀ I r/(2π a²) — zero at the axis, increasing linearly to the surface, then decreasing as 1/r (Fig. 4.14).
4.7 The solenoid
- A solenoid is a long helical coil of many closely spaced turns (Fig. 4.15). Inside a long solenoid the field is uniform, parallel to the axis, and outside it is nearly zero (Fig. 4.16: Amperian rectangular loop abcd: B l = μ₀ (nl) I). B = μ₀ n I, where n is the number of turns per unit length; at the end of the solenoid it is half this value. Using an iron core greatly increases B.
- Example 4.8: solenoid of length 0.5 m, 500 turns, 5 A: n = 1000 turns/m, B = 4π × 10⁻⁷ × 1000 × 5 = 6.28 × 10⁻³ T.
4.8 Force between two parallel currents, the ampere
- Two long parallel wires with currents I₁ and I₂ at separation d (Fig. 4.17): the field of wire 1 at wire 2 is B₁ = μ₀I₁/2πd, so the force per unit length on wire 2 is f = μ₀ I₁ I₂ / (2π d). Parallel currents attract; antiparallel currents repel (Newton’s third law holds: equal and opposite forces).
- The ampere (SI definition, older): 1 A is the steady current which, flowing in two infinitely long, straight, parallel conductors of negligible cross-section placed 1 m apart in vacuum, produces a force of 2 × 10⁻⁷ N per metre of length (the current definition leads to the coulomb as 1 A s). Example 4.9: a very long conductor with 1 A on a table in a place where the horizontal component of Earth’s field is 3.0 × 10⁻⁵ T (south→north): force per unit length = 3 × 10⁻⁵ N m⁻¹ for current east→west, directed downwards (this exceeds the 2 × 10⁻⁷ N m⁻¹ of the ampere definition, so stray fields must be eliminated when standardising), and zero for current along south→north (parallel to B).
4.9 Torque on a current loop, magnetic dipole
4.9.1 Torque on a rectangular loop in a uniform field (Fig. 4.18)
Forces on opposite sides are equal and opposite → net force zero, but forces on the pair of sides ⟂ B form a couple: τ = I A B sin θ (θ between the normal to the loop and B), i.e. τ = m × B with magnetic moment m = N I A (N turns; direction by right-hand rule normal to the plane). Maximum torque when the plane of the loop is parallel to B; zero when the normal is parallel/antiparallel to B (stable equilibrium when m ∥ B). The result is valid for a planar loop of any shape.4.9.2 Circular current loop as a magnetic dipole
At large distances the field of a current loop equals that of a bar magnet/magnetic dipole of moment m = IA (Fig. 4.19); the axial field B = μ₀ 2m/(4π x³); the field lines are similar to those of a bar magnet. Potential energy in a field: U = −m·B.- Example 4.10: 100-turn coil, radius 10 cm, I = 3.2 A: B at centre = 2.0 × 10⁻³ T; magnetic moment m = NIA = 10 A m² (≈ 10.05); in a 2 T field the torque is zero initially (axis along B) and 20 N m (≈ 20.1) after rotating by 90°. Example 4.11: (a) a uniform field cannot rotate a loop about its own axis (the torque is perpendicular to m); (b) stable equilibrium when the plane of the loop is perpendicular to B (m ∥ B), where the flux through the loop is maximum.
4.10 The moving coil galvanometer
- A device to detect and measure small currents (Fig. 4.20): a rectangular coil of N turns and area A on a soft-iron core in the radial magnetic field of curved pole pieces (Fig. 4.21) so that the plane of the coil is always parallel to B and the torque is τ = N I A B for all deflections. This is balanced by the restoring torque of the spring k φ ⇒ φ = (N A B / k) I — deflection proportional to current (a linear scale). Current sensitivity φ/I = NAB/k; voltage sensitivity φ/V = (NAB/k)(1/R). Increasing N increases current sensitivity but also the resistance R, so the voltage sensitivity may not increase.
- Conversion (Fig. 4.22): to an ammeter by a small shunt resistance r_s in parallel (r_s = I_g R_G/(I − I_g)); to a voltmeter by a large series resistance (R = V/I_g − R_G). Example 4.12: compares the current read in the circuit of Fig. 4.23 when the “ammeter” is (a) a galvanometer of resistance 60.00 Ω, (b) the same galvanometer converted to an ammeter by a 0.02 Ω shunt, (c) an ideal ammeter of zero resistance — an ammeter must have a very small resistance, otherwise it alters the very current it measures; the shunted galvanometer reads almost exactly the ideal value.
Summary & “Points to ponder” (NCERT)
- Lorentz force F = q(v × B + E); magnetic force does no work; F = Il × B on a wire; circular motion r = mv/qB, cyclotron frequency independent of speed; Biot–Savart law; loop: B = μ₀I/2R at the centre; Ampere’s law; B = μ₀I/2πr for a long wire and μ₀nI inside a solenoid; parallel currents attract; τ = m × B with m = NIA; galvanometer deflection ∝ current, conversion with shunt/series resistance.
- Electrostatic field lines start at positive charges and end at negative charges, whereas magnetic field lines always form closed loops. The chapter’s laws hold for steady currents; for time-varying currents Newton’s third law is valid only if the momentum carried by the electromagnetic field is included. The velocity-dependent Lorentz force is resolved by switching frames — electricity and magnetism are linked phenomena. Ampere’s law is not independent of the Biot–Savart law (compare Gauss’s law and Coulomb’s law).
A circular coil of wire consisting of 100 turns, each of radius 8.0 cm carries a current of 0.40 A. What is the magnitude of the magnetic field B at the centre of the coil?
Concept and context. A current in a wire produces a magnetic field around it (Oersted, 1820), and the field of a current element is given by the Biot–Savart law: dB = (μ₀/4π) I dl sin θ / r². For a circular loop of radius R carrying current I, every element dl is perpendicular to the line joining it to the centre and at the same distance R from the centre, so all the elements contribute fields in the same direction (along the axis, given by the right-hand rule) and add up. Integrating over the circumference (length 2πR) gives B = μ₀ I / (2R) at the centre. For a tightly wound coil of N turns, each turn contributes the same field, so B = μ₀ N I / (2R). The field lines of the loop are closed curves that are nearly straight and parallel to the axis near the centre (NCERT Fig. 4.10).
Given: N = 100 turns; R = 8.0 cm = 0.080 m; I = 0.40 A; μ₀ = 4π × 10⁻⁷ T m A⁻¹.
- Formula: B = μ₀NI/(2R).
- Substitute: B = (4π × 10⁻⁷ × 100 × 0.40)/(2 × 0.080) = (4π × 10⁻⁷ × 40)/0.16.
- Simplify: 40/0.16 = 250, so B = 4π × 10⁻⁷ × 250 = π × 10⁻³ = 3.14 × 10⁻⁴ T.
- B ≈ 3.1 × 10⁻⁴ T (≈ 3 gauss, comparable to the strength of the Earth’s magnetic field).
B = μ₀NI / 2R = 3.1 × 10⁻⁴ T
| Quantity | Value |
|---|---|
| Number of turns N | 100 |
| Radius R | 0.080 m |
| Current I | 0.40 A |
| μ₀NI | 4π × 10⁻⁷ × 40 = 5.03 × 10⁻⁵ T m |
| B = μ₀NI/2R | 3.14 × 10⁻⁴ T |
- Direction: along the axis of the coil, given by the right-hand thumb rule — curl the fingers in the direction of current; the thumb points in the direction of B.
- Why multiply by N? The turns are close together, so they act like N identical loops carrying the same current, each producing the same field at the centre; the fields add.
- Check on assumptions: the formula is exact for the centre of a single loop and valid for a “closely wound” coil whose thickness is much less than the radius.
Conclusion: the magnitude of the magnetic field at the centre of the coil is ≈ 3.1 × 10⁻⁴ T, directed along the axis of the coil according to the right-hand rule.
A long straight wire carries a current of 35 A. What is the magnitude of the field B at a point 20 cm from the wire?
Concept and context. By symmetry the magnetic field of a long straight current-carrying wire has the same magnitude at all points at a given perpendicular distance r, and it is tangential to circles concentric with the wire (NCERT Fig. 4.1: iron filings form concentric circles; direction given by the right-hand thumb rule). Applying Ampere’s circuital law ∮B·dl = μ₀I_enclosed to a circle of radius r around the wire: B × 2πr = μ₀ I, so B = μ₀ I/(2π r). The field is inversely proportional to the distance from the wire; it falls much more slowly than the 1/r² of a point charge or a current element because we have summed over an infinitely long wire.
Given: I = 35 A; r = 20 cm = 0.20 m; μ₀/2π = 2 × 10⁻⁷ T m A⁻¹.
- Formula: B = μ₀ I/(2π r) = (μ₀/2π) × (I/r).
- Substitute: B = 2 × 10⁻⁷ × 35/0.20 = 2 × 10⁻⁷ × 175 = 3.5 × 10⁻⁵ T.
B = 3.5 × 10⁻⁵ T
- Comparison: 3.5 × 10⁻⁵ T (0.35 gauss) is about the same as the Earth’s horizontal magnetic field (3 × 10⁻⁵ T); so a compass near such a wire would be strongly affected, as in Oersted’s experiment.
- Direction: tangential to the circle of radius r through the point, in the sense given by the right-hand thumb rule (thumb along the current, fingers curling in the direction of B).
- Dependence: B ∝ I and B ∝ 1/r; doubling the distance halves the field.
| Distance r | B = 2 × 10⁻⁷ × 35/r |
|---|---|
| 10 cm | 7.0 × 10⁻⁵ T |
| 20 cm | 3.5 × 10⁻⁵ T |
| 40 cm | 1.75 × 10⁻⁵ T |
Conclusion: the magnitude of the magnetic field at 20 cm from the long straight wire carrying 35 A is 3.5 × 10⁻⁵ T.
A long straight wire in the horizontal plane carries a current of 50 A in north to south direction. Give the magnitude and direction of B at a point 2.5 m east of the wire.
Concept and context. The magnetic field of a long straight wire is B = μ₀I/(2πr), directed tangentially along circles concentric with the wire; the sense is given by the right-hand thumb rule (thumb along the current, fingers curled show B). To find the direction at a given point it is easiest to take unit vectors: let x be east, y north and z vertically upward. The direction of B at a point is that of I × r̂, where r̂ is the unit vector from the wire towards the point.
Given: I = 50 A (north to south); r = 2.5 m (point lies east of the wire).
Magnitude
B = μ₀I/(2πr) = 2 × 10⁻⁷ × 50 / 2.5 = 4 × 10⁻⁶ T
Direction (vectors and right-hand rule)
- Choose axes: x = east, y = north, z = vertically up. The current is from north to south, i.e. along −ŷ; the position of the point relative to the wire is along +x̂ (east).
- The field direction is along I × r̂ = (−ŷ) × (x̂) = −(ŷ × x̂) = −(−ẑ) = +ẑ — vertically upward.
- Right-hand rule check: point the right thumb south along the current. Viewed from the north looking south (the current is going away from you) the field lines circle clockwise; the east side is on your left, where a clockwise circulation points upward, while on the west side (your right) it points downward.
- So at 2.5 m east of the wire, B points vertically upward.
| Item | Result |
|---|---|
| Magnitude | 4 × 10⁻⁶ T (0.04 gauss) |
| Direction | vertically upward (+z) |
| On the west side of the wire (2.5 m) | same magnitude, vertically downward |
| Comparison with Earth’s field | about 1/8 of 3 × 10⁻⁵ T |
Conclusion: the magnetic field at the point 2.5 m east of the wire is 4 × 10⁻⁶ T, directed vertically upward.
A horizontal overhead power line carries a current of 90 A in east to west direction. What is the magnitude and direction of the magnetic field due to the current 1.5 m below the line?
Concept and context. The field of a long straight current-carrying conductor at perpendicular distance r is B = μ₀I/(2πr), tangential to the circle through the point around the wire. Because power lines are very long compared with the 1.5 m separation, the infinite-wire formula is an excellent approximation. For the direction use the right-hand thumb rule or the cross product I × r̂, with r̂ the unit vector from the wire to the point of interest.
Given: I = 90 A (east → west); r = 1.5 m (below the line).
Magnitude
B = μ₀I/(2πr) = 2 × 10⁻⁷ × 90 / 1.5 = 1.2 × 10⁻⁵ T
Direction
- Axes: x = east, y = north, z = up. Current is east → west, i.e. along −x̂. The point is directly below the wire, so r̂ = −ẑ.
- Field direction ∝ I × r̂ = (−x̂) × (−ẑ) = x̂ × ẑ = −ŷ (because ẑ × x̂ = ŷ ⇒ x̂ × ẑ = −ŷ), i.e. towards the south.
- Right-hand rule check: thumb pointing west along the current; the fingers curl from the top of the wire towards the north side (above) and towards the south side (below). Hence beneath the wire the field is directed to the south.
| Position relative to the wire | Magnitude | Direction |
|---|---|---|
| 1.5 m below | 1.2 × 10⁻⁵ T | towards the south |
| 1.5 m above | 1.2 × 10⁻⁵ T | towards the north |
| 1.5 m north (same level) | 1.2 × 10⁻⁵ T | vertically downward |
| 1.5 m south (same level) | 1.2 × 10⁻⁵ T | vertically upward |
- This field (0.12 gauss) is about 40 % of the Earth’s horizontal field; a compass held below such a line would be noticeably deflected.
Conclusion: the magnetic field 1.5 m below the power line is 1.2 × 10⁻⁵ T, directed towards the south.
What is the magnitude of magnetic force per unit length on a wire carrying a current of 8 A and making an angle of 30° with the direction of a uniform magnetic field of 0.15 T?
Concept and context. A charge q moving with velocity v in a magnetic field B experiences the Lorentz force F = q v × B. A current I in a wire of length l consists of a large number of such moving charges (n carriers per unit volume, drift velocity v_d, cross-section A with I = n e A v_d); summing the forces on all the carriers in the length l gives the force on a straight conductor: F = I l × B, where l is a vector of magnitude equal to the length and direction along the current. The magnitude is F = I l B sin θ, θ being the angle between the wire (current direction) and the field. The force is perpendicular to both the wire and the field, with direction from the right-hand rule (NCERT Fig. 4.2); it is maximum (IlB) when the wire is perpendicular to the field and zero when it is parallel.
Given: I = 8 A; B = 0.15 T; θ = 30° (sin 30° = 0.5).
- Use F = I l B sin θ and divide by l: F/l = I B sin θ.
- F/l = 8 × 0.15 × sin 30° = 8 × 0.15 × 0.5 = 0.60 N m⁻¹.
F/l = I B sin θ = 0.6 N m⁻¹
| Angle θ | sin θ | Force per unit length |
|---|---|---|
| 0° (wire ∥ B) | 0 | 0 |
| 30° | 0.5 | 0.60 N m⁻¹ |
| 90° (wire ⟂ B) | 1 | 1.2 N m⁻¹ (maximum) |
- Direction: perpendicular to the plane containing the wire and B, given by l × B (right-hand rule: stretch the fingers along the current, curl them toward B; the thumb gives the force).
- Unit check: A × T = A × N A⁻¹ m⁻¹ = N m⁻¹ ✓.
- Remark: the magnetic force on a current-carrying wire arises because the magnetic force on each moving carrier is transferred to the lattice through collisions (the wire as a whole feels the force).
Conclusion: the magnetic force per unit length on the wire is 0.60 N m⁻¹, directed perpendicular to both the wire and the magnetic field.
A 3.0 cm wire carrying a current of 10 A is placed inside a solenoid perpendicular to its axis. The magnetic field inside the solenoid is given to be 0.27 T. What is the magnetic force on the wire?
Concept and context. The field inside a long solenoid is uniform and directed along the axis (B = μ₀nI), and zero outside (NCERT Fig. 4.16). A short straight wire placed inside the solenoid and perpendicular to its axis is therefore perpendicular to the field, so θ = 90° and the force is simply F = I l B. The force on a current element is F = I l × B (Lorentz force on all moving charges), perpendicular to both the wire and the axis.
Given: l = 3.0 cm = 0.030 m; I = 10 A; B = 0.27 T; θ = 90° (wire ⟂ axis, i.e. ⟂ B).
- F = I l B sin θ = I l B (since sin 90° = 1).
- F = 10 × 0.030 × 0.27 = 0.081 N.
- F = 8.1 × 10⁻² N (81 mN), acting perpendicular to both the wire and the axis of the solenoid (direction by the right-hand rule).
F = IlB = 8.1 × 10⁻² N
- Realism: 0.27 T is a strong field (typical of a laboratory electromagnet — a solenoid with n I ≈ 2 × 10⁵ A m⁻¹ in air); so the force on a mere 3 cm of wire carrying 10 A is appreciable: about the weight of 8 g (0.081 N ÷ 9.8 m s⁻² ≈ 8.3 g).
- Why θ = 90°: the field lines are parallel to the solenoid axis; a wire perpendicular to the axis cuts the field lines at a right angle — giving the maximum force.
| Orientation of wire | θ | Force |
|---|---|---|
| ⟂ axis (given) | 90° | 8.1 × 10⁻² N |
| at 30° to axis | 30° | 4.05 × 10⁻² N |
| along the axis | 0° | 0 |
Conclusion: the magnetic force on the wire is 8.1 × 10⁻² N, acting perpendicular to both the wire and the axis of the solenoid.
Two long and parallel straight wires A and B carrying currents of 8.0 A and 5.0 A in the same direction are separated by a distance of 4.0 cm. Estimate the force on a 10 cm section of wire A.
Concept and context. A current-carrying wire produces a magnetic field around it, and a second current in that field experiences a force. For two long parallel wires carrying currents I_a and I_b at separation d, the field of B at the position of A is B_b = μ₀I_b/(2πd) (directed perpendicular to the plane of the wires), so the force on a length L of A is F = I_a L B_b. Hence force per unit length f = μ₀ I_a I_b/(2π d). Parallel currents attract each other and anti-parallel currents repel — the opposite of the electrostatic rule that like charges repel. The forces on the two wires are equal and opposite, consistent with Newton’s third law. This relation is used to define the ampere.
Given: I_a = 8.0 A; I_b = 5.0 A; d = 4.0 cm = 0.040 m; L = 10 cm = 0.10 m; μ₀/2π = 2 × 10⁻⁷ T m A⁻¹.
- Field of B at A: B_b = μ₀I_b/(2πd) = 2 × 10⁻⁷ × 5.0/0.040 = 2.5 × 10⁻⁵ T.
- Force per unit length on A: f = I_a B_b = 8.0 × 2.5 × 10⁻⁵ = 2.0 × 10⁻⁴ N m⁻¹.
- Force on a 10 cm section of A: F = f × L = 2.0 × 10⁻⁴ × 0.10 = 2.0 × 10⁻⁵ N.
- Direction: the currents are in the same direction ⇒ the force on A is attractive, directed towards wire B (perpendicular to A, in the plane of the two wires).
F = μ₀I_aI_bL/2πd = 2.0 × 10⁻⁵ N (attractive)
| Quantity | Value |
|---|---|
| Field of B at A | 2.5 × 10⁻⁵ T |
| Force per unit length | 2.0 × 10⁻⁴ N m⁻¹ |
| Force on 10 cm of A | 2.0 × 10⁻⁵ N |
| Nature | attractive (same direction currents) |
- Newton’s third law: the force on a 10 cm section of B due to A is also 2.0 × 10⁻⁵ N, but directed towards A — equal and opposite.
- Definition of the ampere: if I_a = I_b = 1 A and d = 1 m then f = 2 × 10⁻⁷ N m⁻¹ — the NCERT definition of 1 A.
- Estimate: the question asks to “estimate” because the formula assumes very long wires (the 10 cm section is far from the ends) and neglects wire thickness.
Conclusion: the force on the 10 cm section of wire A is 2.0 × 10⁻⁵ N, attractive (directed towards wire B), because the currents are in the same direction.
A closely wound solenoid 80 cm long has 5 layers of windings of 400 turns each. The diameter of the solenoid is 1.8 cm. If the current carried is 8.0 A, estimate the magnitude of B inside the solenoid near its centre.
Concept and context. A solenoid is a long helical coil of closely spaced turns. Inside a long solenoid the magnetic field is uniform and directed along the axis, and outside it is nearly zero (Fig. 4.16: apply Ampere’s circuital law to the rectangular loop abcd with one side inside along the axis: B l = μ₀ (n l) I). The result is B = μ₀ n I, where n = number of turns per unit length (turns per metre). It is independent of the radius, and holds near the centre of a solenoid whose length is much greater than its diameter (here 80 cm vs 1.8 cm); near the ends the field falls to about half of this value.
Given: length L = 80 cm = 0.80 m; layers = 5, each with 400 turns ⇒ total turns N = 5 × 400 = 2000; diameter 1.8 cm (so L ≫ diameter, the long-solenoid formula applies); I = 8.0 A.
- Turns per unit length: n = N/L = 2000/0.80 = 2.5 × 10³ turns m⁻¹ (2500 per metre).
- Field: B = μ₀ n I = (4π × 10⁻⁷)(2.5 × 10³)(8.0) = 4π × 10⁻⁷ × 2.0 × 10⁴.
- Calculate: B = 8π × 10⁻³ T = 2.51 × 10⁻² T ≈ 2.5 × 10⁻² T.
B = μ₀nI = 2.5 × 10⁻² T
- Remark on layers: the multiple layers all carry current in the same sense, so the total turns per unit length is the sum over layers; the layer radii differ slightly (diameter 1.8 cm), but the field depends on n only and not on the radius.
- Direction: along the axis, given by the right-hand rule (curl the fingers along the current in the turns; the thumb points along B).
- Iron core: inserting a soft-iron core would increase B by a factor equal to its relative permeability — the principle of the electromagnet.
- At the ends of the solenoid B is about half this value (1.3 × 10⁻² T).
| Quantity | Value |
|---|---|
| Total turns N | 2000 |
| n = N/L | 2500 m⁻¹ |
| μ₀nI | 4π × 10⁻⁷ × 2500 × 8 = 2.5 × 10⁻² T |
| Field outside (ideal long solenoid) | ≈ 0 |
Conclusion: the magnetic field inside the solenoid near its centre is ≈ 2.5 × 10⁻² T (8π × 10⁻³ T), directed along the axis.
A square coil of side 10 cm consists of 20 turns and carries a current of 12 A. The coil is suspended vertically and the normal to the plane of the coil makes an angle of 30° with the direction of a uniform horizontal magnetic field of magnitude 0.80 T. What is the magnitude of torque experienced by the coil?
Concept and context. A current loop in a uniform magnetic field experiences no net force (the forces on opposite sides are equal and opposite) but does experience a torque, because the forces on the two sides perpendicular to the field act along different lines and form a couple (NCERT Fig. 4.18). For a coil of N turns, area A and current I, with the normal to its plane making an angle θ with B, the torque is τ = N I A B sin θ = m B sin θ, i.e. τ = m × B, where m = NIA is the magnetic dipole moment of the coil (direction along the normal given by the right-hand rule). The torque tends to rotate the coil until m aligns with B (the plane of the coil perpendicular to B). It is zero when θ = 0° or 180° and maximum when the plane of the coil is parallel to B (θ = 90°).
Given: side a = 10 cm = 0.10 m ⇒ area A = a² = 1.0 × 10⁻² m²; N = 20 turns; I = 12 A; B = 0.80 T; θ = 30° (angle between normal and B); sin 30° = 0.5.
- Magnetic moment of the coil: m = N I A = 20 × 12 × 1.0 × 10⁻² = 2.4 A m².
- Torque: τ = m B sin θ = 2.4 × 0.80 × 0.5 = 0.96 N m.
- Direction: along m × B — it turns the coil so as to align the normal with the field (reduce θ to zero).
τ = NIAB sin θ = 0.96 N m
| Quantity | Value |
|---|---|
| Area A | 1.0 × 10⁻² m² |
| Magnetic moment m = NIA | 2.4 A m² |
| θ (normal, B) | 30° |
| τ = mB sin θ | 0.96 N m |
- Caution about the angle: in τ = NIAB sin θ, θ is the angle between the normal to the coil and B. If the question had given the angle between the plane of the coil and B (e.g. 30°), then θ would be 60° and sin θ = 0.87.
- Check of limits: maximum torque (plane parallel to B, θ = 90°) would be NIAB = 1.92 N m; minimum torque is zero.
Conclusion: the torque experienced by the coil is 0.96 N m, tending to rotate the coil so that its normal aligns with the magnetic field.
Two moving coil meters, M₁ and M₂ have the following particulars: R₁ = 10 Ω, N₁ = 30, A₁ = 3.6 × 10⁻³ m², B₁ = 0.25 T; R₂ = 14 Ω, N₂ = 42, A₂ = 1.8 × 10⁻³ m², B₂ = 0.50 T (The spring constants are identical for the two meters). Determine the ratio of (a) current sensitivity and (b) voltage sensitivity of M₂ and M₁.
Concept and context. The moving coil galvanometer (NCERT Fig. 4.20) consists of a rectangular coil of N turns and area A, pivoted in the radial magnetic field (Fig. 4.21) between the curved pole pieces of a permanent magnet, with a soft-iron core. The radial field keeps the plane of the coil always parallel to B, so the torque is τ = N I A B for every deflection. A hair spring provides a restoring torque kφ (k = torsion constant). In equilibrium N I A B = k φ, so the deflection φ = (NAB/k) I is proportional to the current (a linear scale). Current sensitivity = φ/I = NAB/k (deflection per unit current). Voltage sensitivity = φ/V = (NAB/k)(1/R), deflection per unit voltage (with V = IR, R = resistance of the galvanometer coil).
Data: M₁: R₁ = 10 Ω, N₁ = 30, A₁ = 3.6 × 10⁻³ m², B₁ = 0.25 T. M₂: R₂ = 14 Ω, N₂ = 42, A₂ = 1.8 × 10⁻³ m², B₂ = 0.50 T. Same spring constant k.
(a) Ratio of current sensitivities
- Current sensitivity S_I = NAB/k. For identical k: S_I2/S_I1 = (N₂A₂B₂)/(N₁A₁B₁).
- N₁A₁B₁ = 30 × 3.6 × 10⁻³ × 0.25 = 2.7 × 10⁻² ; N₂A₂B₂ = 42 × 1.8 × 10⁻³ × 0.50 = 3.78 × 10⁻².
- Ratio = 3.78 × 10⁻² / 2.7 × 10⁻² = 1.4.
(b) Ratio of voltage sensitivities
- Voltage sensitivity S_V = S_I/R = NAB/(kR).
- S_V2/S_V1 = (S_I2/S_I1) × (R₁/R₂) = 1.4 × (10/14) = 1.4 × 0.714 = 1.0.
SI2/SI1 = 1.4 ; SV2/SV1 = 1.0
| Quantity | M₁ | M₂ | Ratio M₂/M₁ |
|---|---|---|---|
| N A B | 2.7 × 10⁻² | 3.78 × 10⁻² | 1.4 (current sensitivity) |
| Resistance R | 10 Ω | 14 Ω | 1.4 |
| NAB/R | 2.7 × 10⁻³ | 2.7 × 10⁻³ | 1.0 (voltage sensitivity) |
- Interpretation: M₂ is 40 % more sensitive to current, but because its coil has more turns its resistance is also 40 % higher, so its voltage sensitivity is the same as M₁. In general, increasing the number of turns N raises current sensitivity (∝ N) but also raises R (∝ N), so voltage sensitivity (∝ N/R) does not improve.
- Ways to increase sensitivity: larger N, A, B; smaller k (weaker spring); the radial field gives linear scale.
Conclusion: the ratio of current sensitivity of M₂ to M₁ is 1.4, and the ratio of their voltage sensitivities is 1 (equal), since the higher current sensitivity of M₂ is offset by its proportionally higher resistance.
In a chamber, a uniform magnetic field of 6.5 G (1 G = 10⁻⁴ T) is maintained. An electron is shot into the field with a speed of 4.8 × 10⁶ m s⁻¹ normal to the field. Explain why the path of the electron is a circle. Determine the radius of the circular orbit. (e = 1.5 × 10⁻¹⁹ C, mₑ = 9.1 × 10⁻³¹ kg)
Concept and context. A charge q moving with velocity v in a magnetic field B experiences the Lorentz (magnetic) force F = q (v × B). This force is always perpendicular to the velocity (and to B). Therefore it does no work on the particle — the speed (kinetic energy) remains constant while only the direction of velocity changes. If the velocity is exactly normal to B the force has constant magnitude qvB and always lies in the plane perpendicular to B, pointing toward a fixed centre: it acts as a centripetal force, and the particle moves in a circle (NCERT Fig. 4.5). If the velocity had a component along B, that component would be unaffected and the path would be a helix (Fig. 4.6).
(i) Why the path is circular (explanation)
- The magnetic force F = −e(v × B) is perpendicular to v at every instant, so it cannot change the speed; the kinetic energy of the electron is constant: |v| = 4.8 × 10⁶ m s⁻¹.
- Because v ⟂ B, the magnitude of the force is F = evB — constant — and its direction is always perpendicular to the velocity and in the plane normal to B. A force of constant magnitude that is always perpendicular to the velocity is exactly a centripetal force; it makes the particle describe a uniform circular motion in the plane perpendicular to B.
- Equating the magnetic force with the centripetal force: evB = mv²/r ⇒ r = mv/(eB) (radius proportional to momentum).
(ii) Radius of the orbit — Given: B = 6.5 G = 6.5 × 10⁻⁴ T; v = 4.8 × 10⁶ m s⁻¹; e = 1.5 × 10⁻¹⁹ C (as given in the question); mₑ = 9.1 × 10⁻³¹ kg.
- r = mv/(eB) = (9.1 × 10⁻³¹ × 4.8 × 10⁶)/(1.5 × 10⁻¹⁹ × 6.5 × 10⁻⁴).
- Numerator = 4.368 × 10⁻²⁴; denominator = 9.75 × 10⁻²³.
- r = 4.48 × 10⁻² m ≈ 4.5 cm.
r = mv/eB ≈ 4.5 × 10⁻² m (4.5 cm) with the value e = 1.5 × 10⁻¹⁹ C given in the question
- Note on the value of e: the question (as printed in this NCERT edition) gives e = 1.5 × 10⁻¹⁹ C. With the standard value e = 1.6 × 10⁻¹⁹ C the radius is r = 4.368 × 10⁻²⁴/(1.6 × 10⁻¹⁹ × 6.5 × 10⁻⁴) = 4.2 cm (the value found with the accepted constant). Use the value given in the question; both are about 4–5 cm.
- Sense of rotation: with B directed into the page, an electron at the top of its orbit moving to the right feels a force F = −e(v × B) pointing downward (towards the centre), so it circulates clockwise; a positive charge in the same field would circulate anticlockwise.
| Quantity | Value |
|---|---|
| m v | 4.37 × 10⁻²⁴ kg m s⁻¹ |
| e B | 9.75 × 10⁻²³ C T |
| r = mv/eB | 4.5 cm (4.2 cm with e = 1.6 × 10⁻¹⁹ C) |
| Kinetic energy | ½mv² = 1.05 × 10⁻¹⁷ J ≈ 65 eV (constant) |
Conclusion: the electron describes a circle because the magnetic force is always perpendicular to its velocity and supplies the centripetal force; the radius of the orbit is r = mv/eB ≈ 4.5 cm (≈ 4.2 cm with e = 1.6 × 10⁻¹⁹ C).
In Exercise 4.11 obtain the frequency of revolution of the electron in its circular orbit. Does the answer depend on the speed of the electron? Explain.
Concept and context. For a charged particle in a uniform magnetic field with v ⟂ B, the magnetic force provides the centripetal force: qvB = mv²/r, giving r = mv/(qB). The angular frequency is ω = v/r = qB/m, and the frequency of revolution (the cyclotron frequency) is ν = ω/2π = qB/(2πm). Notice that it depends only on the charge-to-mass ratio q/m of the particle and on the field B — not on the speed or the radius. This independence is exploited in the cyclotron to accelerate charged particles with an alternating electric field of fixed frequency.
Given (from Exercise 4.11): B = 6.5 × 10⁻⁴ T; e = 1.5 × 10⁻¹⁹ C; mₑ = 9.1 × 10⁻³¹ kg.
- ν = eB/(2π mₑ) = (1.5 × 10⁻¹⁹ × 6.5 × 10⁻⁴)/(2 × 3.14 × 9.1 × 10⁻³¹).
- Numerator = 9.75 × 10⁻²³; denominator = 5.72 × 10⁻³⁰.
- ν ≈ 1.7 × 10⁷ Hz = 17 MHz (with e = 1.6 × 10⁻¹⁹ C it is 1.8 × 10⁷ Hz ≈ 18 MHz).
- Time period T = 1/ν ≈ 5.9 × 10⁻⁸ s (59 ns).
ν = eB/2πm ≈ 1.7 × 10⁷ Hz
Does the frequency depend on the speed?
- No. The expression ν = eB/(2πm) contains only e, m and B. A faster electron has a larger radius (r = mv/eB ∝ v) but it also has to travel a proportionally longer circumference (2πr ∝ v) — the two effects cancel exactly, so the time for one revolution is the same: T = 2πr/v = 2πm/eB.
- Consequence: electrons (or ions) of different speeds injected into the same field circulate with the same frequency; slower ones on small circles and faster ones on big circles. This is the principle of the cyclotron: the particles are accelerated across a gap by an oscillating electric field of the same frequency, spiral outward gaining energy and always arriving at the gap in step.
- Limit: at speeds close to that of light the relativistic increase of mass makes the frequency decrease slightly; here v/c = 1.6 × 10⁻² so the non-relativistic treatment is valid.
| Speed v | Radius r = mv/eB | Frequency ν |
|---|---|---|
| v | r | 1.7 × 10⁷ Hz |
| 2v | 2r | 1.7 × 10⁷ Hz (same) |
| v/2 | r/2 | 1.7 × 10⁷ Hz (same) |
Conclusion: the frequency of revolution of the electron is ν = eB/2πm ≈ 1.7 × 10⁷ Hz (≈ 17 MHz) and it does not depend on the speed of the electron — a faster electron moves on a proportionally larger circle in the same time.
(a) A circular coil of 30 turns and radius 8.0 cm carrying a current of 6.0 A is suspended vertically in a uniform horizontal magnetic field of magnitude 1.0 T. The field lines make an angle of 60° with the normal of the coil. Calculate the magnitude of the counter torque that must be applied to prevent the coil from turning. (b) Would your answer change, if the circular coil in (a) were replaced by a planar coil of some irregular shape that encloses the same area? (All other particulars are also unaltered.)
Concept and context. A planar current loop of area A with N turns and current I has a magnetic dipole moment m = NIA (directed along the normal by the right-hand rule). In a uniform field it experiences the torque τ = m × B, of magnitude NIAB sin θ, where θ is the angle between the normal (m) and B. To keep the coil from turning, an equal and opposite counter torque must be applied (e.g. by the suspension fibre or a spring).
Given: N = 30; R = 8.0 cm = 0.080 m ⇒ A = πR² = 3.14 × (0.080)² = 2.01 × 10⁻² m²; I = 6.0 A; B = 1.0 T; θ = 60° between the field lines and the normal; sin 60° = 0.866.
(a) Counter torque
- Magnetic moment: m = NIA = 30 × 6.0 × 2.01 × 10⁻² = 3.62 A m².
- Torque: τ = m B sin θ = 3.62 × 1.0 × 0.866 = 3.13 N m.
- The counter torque must have the same magnitude: ≈ 3.1 N m, in the sense opposite to that of the magnetic torque (which tends to turn the coil so that its normal aligns with B).
τ = NIAB sin θ = 3.1 N m
(b) Irregular planar coil of the same area
- No, the answer would not change. The torque on a planar current loop in a uniform magnetic field, τ = NIAB sin θ, depends only on the area vector A (its magnitude and orientation), the current and the field. It is independent of the shape of the loop — for any planar shape the loop can be regarded as made up of many small rectangular current loops (the currents in the interior edges cancel), each experiencing the torque I dA × B, and these add up to I A × B.
- Since the area (hence the magnetic moment m = NIA), the current, the field and the orientation are all unchanged, the torque and so the counter torque remain 3.1 N m.
| Case | Area A | m = NIA | τ = mB sin 60° |
|---|---|---|---|
| Circular coil (given) | 2.01 × 10⁻² m² | 3.62 A m² | 3.13 N m |
| Irregular planar coil, same area | same | same | same (3.13 N m) |
Conclusion: (a) the counter torque required is ≈ 3.1 N m; (b) the answer would not change for an irregular planar coil of the same area, since the torque depends only on the area, not on the shape.