Moving charges and magnetism | ISC Class 12 Physics Notes

This note covers magnetic fields due to currents, magnetic forces on charges and conductors, particle motion, the Biot-Savart and Ampere laws, circular coils and solenoids, forces between parallel wires, magnetic torque and dipole moment, and moving coil galvanometers with ammeter and voltmeter conversions.

How does a current produce a magnetic field?

A magnetic field describes the magnetic influence at a point in space. It is a vector quantity, meaning that it has both magnitude and direction. An electric current is the rate of charge flow. Moving charges and electric currents produce magnetic fields in the surrounding space.

What did Oersted observe?

In 1820, Oersted observed that a current in a straight wire deflected a nearby magnetic compass needle. A compass needle is a small magnet free to turn. Its alignment near the wire was tangential to a circle centred on the wire.

The circular alignment is noticeable when the current is large and the needle sufficiently close to the wire for the Earth's magnetic field to be ignored. Increasing the current or bringing the needle closer increases its deflection. Reversing the current reverses the needle's orientation.

Iron filings around the wire arrange themselves in concentric circles. These observations connect current with magnetism: the wire produces a magnetic field, which acts on the compass needle.

Two rings of compass needles surround a central dot and a central cross respectively. The dark ends mark north poles. A third panel shows concentric arrangements of iron filings around the wire.

See Fig. 4.1 in your NCERT textbook

How are directions and combined fields represented?

A dot represents a direction out of the page; a cross represents a direction into it. For a straight wire, point the right thumb along conventional current, the direction of positive charge flow. Curled fingers show the magnetic field direction.

The principle of superposition means that fields from several sources add as vectors. Magnetic field lines form closed loops. Electrostatic field lines originate at positive charges and terminate at negative charges or fade at infinity. Magnetic monopoles, isolated magnetic poles, are not known to exist.

What force acts on a moving charge in electric and magnetic fields?

Let q be a particle's signed charge, v its velocity vector, E the electric field vector, and B the magnetic field vector at its position. Electric field is electric force per unit test charge. Their magnitudes are written without arrows. An arrow over a symbol explicitly indicates a vector.

The Lorentz force, the total electric and magnetic force F⃗, is:

F⃗ = q(E⃗ + v⃗ × B⃗)

The multiplication sign between vectors denotes a cross product: its magnitude is the product of their magnitudes and the sine of their included angle; its direction is perpendicular to both. The magnetic part alone is F⃗ᵦ = q(v⃗ × B⃗).

Define Fᵦ as the magnitude of magnetic force and θ as the angle between velocity and field. Then Fᵦ = |q|vB sinθ, where |q| means the positive magnitude of charge. Curl the right-hand fingers from velocity towards field through the smaller angle; the thumb gives the force direction for positive charge. Reverse it for negative charge.

Which special cases matter?

The magnetic force is perpendicular to velocity, so it does no work. It changes the direction of motion without changing speed. The electric force qE⃗ can have a component along motion and can therefore transfer energy.

Definition: A field has magnitude one tesla when a charge of one coulomb moving perpendicular to it at one metre per second experiences a magnetic force of one newton.

The SI unit of magnetic field is tesla, symbol T. The other unit symbols here are N for newton, s for second, C for coulomb and m for metre: 1 T = 1 N s/(C m). Magnetic field is defined through force on a moving charge.

Mass is measured in kilograms (kg); one gram (g) is 10⁻³ kg. One centimetre (cm) is 10⁻² m. These conversions put the supplied lengths and masses into SI units before substitution.

How does a charged particle move in a uniform magnetic field?

A uniform field has the same magnitude and direction throughout the region considered. When magnetic force is the only force and velocity is perpendicular to the field, the force continually turns the particle towards the centre of a circular path.

Derivation: Radius and period of circular motion

Let m be the particle's mass, r the radius of its path and v its speed. Centripetal force means the inward force required to maintain circular motion. Use the non-relativistic model, in which the particle's mass is treated as constant.

1. The centripetal force required is mv²/r. Magnetic force has magnitude |q|vB because velocity and field are perpendicular.
2. Equate the two forces: mv²/r = |q|vB. Cancelling v gives the radius r = mv/(|q|B).
3. Let Tₚ be the period, the time for one revolution. Dividing the circumference by speed gives Tₚ = 2πr/v = 2πm/(|q|B), where π is the circle constant.
4. Let f be the frequency, the number of revolutions per second. Since f = 1/Tₚ, the frequency is |q|B/(2πm).

r = mv/(|q|B)

f = |q|B/(2πm)

The SI unit of frequency is hertz, symbol Hz, equal to one cycle per second. In this model, frequency is independent of speed. A greater momentum mv gives a larger radius for a fixed charge and field.

If velocity also has a component along the field, that component remains unchanged. Circular motion across the field combines with uniform motion along it to form a helix, a winding path around an axis. Its pitch is the distance travelled along the field in one revolution.

Worked example 1. An electron of mass 9 × 10⁻³¹ kg and charge magnitude 1.6 × 10⁻¹⁹ C moves at 3 × 10⁷ m/s perpendicular to a field of 6 × 10⁻⁴ T. Find its orbital radius and frequency.

Answer: Formula: r = mv/(|q|B); f = v/(2πr). Substitute: r = (9 × 10⁻³¹ × 3 × 10⁷)/(1.6 × 10⁻¹⁹ × 6 × 10⁻⁴) ≈ 0.28 m. Then f ≈ 17 × 10⁶ Hz. The negative charge affects the direction of rotation, not these magnitudes.

How does the Biot-Savart law describe the field of a current element?

A current element is a very short directed length of a current-carrying conductor. Let I be the steady current and dℓ⃗ the element, directed along current. Let r⃗ point from the element to the observation point, with magnitude r and unit direction r̂.

A unit vector has magnitude one. Let dB⃗ denote the small field contribution of the element and μ₀ the permeability of free space, the constant relating current to magnetic field in vacuum. The Biot-Savart law gives:

dB⃗ = (μ₀/4π)I(dℓ⃗ × r̂)/r²

With θ now denoting the angle between dℓ⃗ and r⃗, its magnitude is dB = μ₀I dℓ sinθ/(4πr²). The field is perpendicular to the plane of these two vectors. Contributions from the whole conductor must be added vectorially.

What conditions and units accompany the law?

This expression applies to steady currents in vacuum. Steady means unchanging with time. For the calculations here, use μ₀ = 4π × 10⁻⁷ T m A⁻¹, where A denotes ampere, the unit of current. The SI unit of permeability is T m A⁻¹.

The field contribution increases with current and element length. For a fixed angle, it decreases with the square of distance. A point along the current element's line has zero contribution from that element because the cross product vanishes.

Worked example 2. A short 1 cm element carries 10 A along the positive x-axis at the origin. Find its field contribution at a point 0.5 m along the positive y-axis, using μ₀/(4π) = 10⁻⁷ T m A⁻¹. The x-, y- and z-axes form a right-handed coordinate system.

Answer: The length is 0.01 m and the included angle is 90°. Substitution gives dB ≈ 10⁻⁷ × 10 × 0.01/(0.5)² = 4 × 10⁻⁸ T. The cross product of the positive x and positive y directions points along the positive z-axis.

What is the finite straight-wire formula?

For a straight wire, let a be the perpendicular distance from the observation point to the wire. If the perpendicular foot lies between the ends, let α and β be the positive angles subtended by the ends relative to that perpendicular.

B = μ₀I(sinα + sinβ)/(4πa)

This finite-wire result requires the stated angle convention. The field direction follows the right-hand rule. For an infinitely long wire both angles become 90°, giving B = μ₀I/(2πa).

How is the magnetic field of a circular coil obtained?

Consider a circular loop of radius R carrying steady current I in vacuum. Its axis is the line through its centre perpendicular to its plane. Let x be the distance along that axis from the centre to an observation point.

Derivation: Field on the axis and at the centre

1. Each current element is the same distance r = √(R² + x²) from the axial point. Its direction is perpendicular to the displacement towards that point.
2. Biot-Savart therefore gives dB = μ₀I dℓ/[4π(R² + x²)]. Resolve this small field into components along and perpendicular to the axis.
3. Perpendicular components from diametrically opposite elements cancel. Each axial component equals dB multiplied by R/√(R² + x²), so these components reinforce one another.
4. Adding all elements replaces their total length by the circumference 2πR. For N closely wound identical turns, multiply the single-loop field by N.

B = μ₀NIR²/[2(R² + x²)³ᐟ²]

Here N is the number of turns. At the centre, x = 0, so B = μ₀NI/(2R). The centre expression also follows directly by adding the equal, similarly directed field contributions of all elements around the loop.

A loop centred at O lies in the YZ plane. The point P lies on the X-axis. The diagram labels radius R, axial distance x, displacement r and the axial and perpendicular components of the small field dB.

See Fig. 4.9 in your NCERT textbook

How is the field direction found?

Curl the right-hand fingers in the direction of current around the coil. The extended thumb gives the axial field direction. Reversing current reverses this direction. For a tightly wound coil, treating all turns as having the same radius permits their fields to be multiplied by N.

Worked example 3. A tightly wound coil has 100 turns, radius 10 cm and current 1 A. Calculate its central field using μ₀ = 4π × 10⁻⁷ T m A⁻¹.

Answer: Convert the radius to 0.1 m. With N = 100 and I = 1 A, B = μ₀NI/(2R) = (4π × 10⁻⁷ × 100 × 1)/(2 × 0.1) = 6.28 × 10⁻⁴ T.

How is Ampere's circuital law applied to a wire and a solenoid?

Ampere's circuital law relates the magnetic field around a closed path to the net current passing through a surface bounded by that path. For steady currents in vacuum:

∮ B⃗ · dℓ⃗ = μ₀Iₑ

The symbol ∮ means integration around a closed path. The dot denotes a scalar product, here the tangential field component multiplied by the small path length. The symbol Iₑ is the algebraic current enclosed, with signs set by the right-hand rule.

Curl the right-hand fingers in the direction of traversal of the path. The thumb marks positive current through its surface. The chosen closed path is an Amperian loop. Symmetry makes the integral easy to evaluate in suitable arrangements.

Derivation: Field near an infinitely long straight wire

1. Choose a circular Amperian loop of radius r centred on the straight wire in a plane perpendicular to it. The observation points are outside the wire.
2. By cylindrical symmetry, the field is tangential everywhere on the circle and has constant magnitude B along it.
3. The integral therefore becomes B multiplied by the circumference: B(2πr) = μ₀I, because the full wire current is enclosed.

B = μ₀I/(2πr)

Thus the field weakens inversely with distance from the wire and increases directly with current. Ampere's law holds for any closed path under the stated steady-current conditions, but a poorly chosen path need not make the field calculable easily.

What is the field inside a long solenoid?

A solenoid is a wire wound into closely spaced helical turns. A long solenoid has length large compared with its radius. The fields of its turns reinforce inside, giving a strong field approximately parallel to the axis near the middle.

The exterior field approaches zero as the solenoid becomes very long. In the ideal long-solenoid treatment it is taken as zero. A rectangular Amperian path has one long side inside and one outside; the transverse sides contribute no tangential field.

If n is the number of turns per unit length, Ampere's law gives B = μ₀nI inside the ideal long solenoid in vacuum. This is a qualitative field model: a finite solenoid has a weak exterior field and its end regions differ from its middle.

What force acts on a current-carrying conductor?

Current consists of moving charges, so an external magnetic field can exert a force on a conductor. Let dF⃗ be the force on a small current element. In the charge-force relation, write velocity as dℓ⃗/dt and current as dq/dt.

Here dt is a small time interval and dq is the charge passing in that interval. The result is dF⃗ = I(dℓ⃗ × B⃗). For a straight length ℓ in a uniform external field, adding the elements gives F⃗ = I(ℓ⃗ × B⃗).

The vector ℓ⃗ points along conventional current and has magnitude ℓ. If θ is its angle with the magnetic field, the force magnitude is F = IℓB sinθ. The right-hand cross-product rule gives its direction.

Which field belongs in the equation?

Note: Use the external magnetic field acting on the conductor, not the field created by that same conductor. Current I is a scalar, a quantity without a vector direction; the direction in the cross product is supplied by the length vector.

A conductor parallel to the field experiences no magnetic force from it. For a perpendicular conductor the magnitude is IℓB. For a curved wire, consider short straight elements and add their forces as vectors rather than treating its total length as one directed segment.

Worked example 4. A straight horizontal wire of mass 200 g and length 1.5 m carries 2 A. A uniform horizontal field perpendicular to the wire supports it against gravity. Find the field magnitude, using gravitational acceleration g = 9.8 m/s² and neglecting the Earth's field.

Answer: Formula: F = IℓB; B = mg/(Iℓ). Substitute: m = 0.2 kg, so B = (0.2 × 9.8)/(2 × 1.5) ≈ 0.65 T. The field must be directed so that the magnetic force is upward and balances the downward weight mg.

Why do parallel currents attract, and how is the ampere related to this force?

Consider two long straight parallel wires in vacuum separated by distance d. Let their currents be I₁ and I₂. The first wire creates an external field at the second, and that field acts on the second wire's current.

Derivation: Force between parallel wires

1. The field at the second wire due to the first has magnitude B₁ = μ₀I₁/(2πd), where B₁ labels that field.
2. For a length L of the second wire, the field is perpendicular to current. Its magnetic force magnitude is F = I₂LB₁.
3. Substituting the field gives F = μ₀I₁I₂L/(2πd). Dividing by L gives the force per unit length.
4. The right-hand rules show attraction for currents in the same direction and repulsion for currents in opposite directions. Forces on corresponding lengths are equal and opposite.

F/L = μ₀I₁I₂/(2πd)

This result is consistent with Newton's third law for these parallel conductors and steady currents. Do not replace the current-direction test with the electrostatic rule for like charges: the interactions being compared are different.

What is the force-based definition of the ampere?

In the historical force-based definition, one ampere is the steady current which, maintained in each of two very long straight parallel conductors of negligible cross-section, one metre apart in vacuum, produces a force of 2 × 10⁻⁷ newtons per metre on each conductor.

The SI unit of current is ampere. In applying the force comparison, the wire spacing, vacuum condition and negligible cross-section all matter. Stray magnetic fields must be eliminated because they can add forces unrelated to the other wire's current.

Let Q be charge transferred in time t by steady current I. Then Q = It. The SI unit of charge is coulomb: 1 C = 1 A s, the charge passing a cross-section in one second when the current is one ampere.

Why does a current loop experience torque in a uniform field?

A closed current loop in a uniform magnetic field has zero resultant magnetic force, but can experience torque, the turning effect of forces. Equal and opposite forces can act along different lines, forming a couple which rotates the loop.

Derivation: Torque on a rectangular coil

Let the rectangular sides have lengths a and b, and let A = ab be the area of one turn. Choose the sides of length b perpendicular to the magnetic field, with the rotation axis parallel to them and passing through the midpoints of the sides of length a. Let θ be the angle between the field and the normal, a direction perpendicular to the coil's plane.

1. The forces on one opposite pair of sides cancel without producing a turning effect about the coil's rotation axis.
2. The other pair experiences equal and opposite forces of magnitude IbB. Their lines of action are separated by perpendicular distance a sinθ.
3. The torque magnitude is force multiplied by this separation: τ = IbB(a sinθ) = IAB sinθ, where τ denotes torque.
4. For N closely wound turns, the torques add, giving τ = NIAB sinθ.

τ = NIAB sinθ

The SI unit of torque is newton metre, written N m. The angle belongs to the normal, not the plane: torque is greatest when the plane is parallel to the field and zero when its normal is parallel or antiparallel to the field.

The formula uses the external field. The field generated by the coil at its centre is a separate quantity. Confusing the two fields changes both the physical meaning and the numerical result.

Worked example 5. A closely wound circular coil has 100 turns, radius 10 cm and current 3.2 A. Find its magnetic moment and the torque magnitude when its normal is perpendicular to an external field of 2 T. Use π = 3.14.

Answer: The magnetic moment mᵈ is current multiplied by total turn-area. Formula: A = πR²; mᵈ = NIA; τ = mᵈB sinθ. Substitute: R = 0.1 m, so A = 3.14 × 10⁻² m², mᵈ ≈ 10 A m² and τ ≈ 10 × 2 = 20 N m.

How does a current loop act as a magnetic dipole?

A magnetic dipole has a magnetic moment which describes its strength and orientation. A current loop behaves as a magnetic dipole at distances large compared with its radius, and experiences a torque like a magnetic needle in an external field.

Write its magnetic moment as m⃗ᵈ to distinguish it from particle mass m. Define the area vector A⃗ as a vector perpendicular to the loop, with magnitude equal to its area and direction fixed by the right-hand current rule.

m⃗ᵈ = NIA⃗

The SI unit of magnetic moment is ampere square metre, written A m². Curling the right-hand fingers along current gives the direction of both A⃗ and m⃗ᵈ through the extended thumb.

Which orientation is stable?

The vector torque relation is τ⃗ = m⃗ᵈ × B⃗. Parallel moment and field give stable equilibrium: a small displacement produces a restoring torque. Antiparallel moment and field give unstable equilibrium: a small displacement produces a torque that moves the loop further away.

Both positions have zero torque exactly at equilibrium. Therefore, zero torque alone does not establish stability. The response to a small angular displacement is the deciding test.

What orbital magnetic moment does a revolving electron have?

In the Bohr model of the hydrogen atom, an electron revolving around the nucleus constitutes a current loop. Let e be the positive magnitude of electron charge, mₑ its mass, and L⃗ its orbital angular momentum, the rotational momentum associated with its orbit.

The electron's orbital magnetic moment is m⃗ᵈ = −(e/2mₑ)L⃗. The minus sign means that its magnetic moment is opposite to its orbital angular momentum because the electron is negatively charged. This orbital contribution is distinct from the electron's intrinsic magnetic moment.

How does a moving coil galvanometer work, and what determines its sensitivity?

A moving coil galvanometer detects small currents by the deflection of a current-carrying coil in a magnetic field. Its many-turn coil can rotate about a fixed axis. A spring supplies a restoring torque, and a pointer indicates deflection on a scale.

A cylindrical soft iron core strengthens the magnetic field and helps make it radial. A radial field is arranged so that the field lies in the coil's plane as it turns. The angle between field and coil normal therefore remains 90°.

The drawing labels a scale, pointer, coil, spring, pivot and soft-iron core between the N and S poles of a permanent magnet. Radial lines surround the core, and the lower label identifies the uniform radial magnetic field.

See Fig. 4.20 in your NCERT textbook

How is deflection related to current?

Let φ be the angular deflection and Cₜ the spring's torsional constant, its restoring torque per unit angular twist. The magnetic torque is NIAB. At equilibrium it balances the spring torque: Cₜφ = NIAB.

Hence φ = NIAB/Cₜ. Equivalently, I = kφ, where k = Cₜ/(NAB) is the current required per unit deflection. For a given galvanometer these quantities are constant, so deflection is proportional to current and permits a uniform scale.

Current sensitivity Sᵢ means deflection per unit current: Sᵢ = φ/I = NAB/Cₜ. It increases with number of turns, coil area or field strength, and increases when the torsional constant is reduced, other factors remaining the same.

Why is voltage sensitivity different?

Let V be the voltage, or potential difference, across the coil and Rɢ its resistance, the ratio of voltage to current. Potential difference is energy transferred per unit charge; its unit is the volt, also written V. Voltage sensitivity Sᵥ means deflection per unit voltage. Since V = IRɢ, Sᵥ = φ/V = NAB/(CₜRɢ).

Increasing current sensitivity may not necessarily increase voltage sensitivity. Doubling the turns doubles current sensitivity if other factors remain constant, but resistance is also likely to double because wire length increases. If it doubles, the voltage sensitivity remains unchanged.

How is a galvanometer converted into an ammeter or a voltmeter?

An ammeter measures current and is connected in series with the circuit. A voltmeter measures potential difference and is connected in parallel across the relevant component. Their resistances must limit the disturbance caused by inserting the measuring instrument.

How does a shunt produce an ammeter?

A shunt is a small resistance connected in parallel with the galvanometer, allowing most of the current to bypass its coil. Let Iɢ be the galvanometer's full-scale current, the current giving maximum scale deflection, and I the desired ammeter range.

If S is shunt resistance, equal potential differences across the parallel branches give IɢRɢ = (I − Iɢ)S. Therefore S = IɢRɢ/(I − Iɢ). The combined resistance is RɢS/(Rɢ + S), approximately S when S is much smaller than Rɢ.

How does a series resistance produce a voltmeter?

For a desired full-scale voltage V, add resistance Rₛ in series with the galvanometer. At full scale, V = Iɢ(Rɢ + Rₛ), so Rₛ = V/Iɢ − Rɢ. The complete instrument has high resistance and draws a very small current.

The ammeter circuit places a shunt branch below and in parallel with the galvanometer branch. The voltmeter circuit adds a resistor in the same series path as the galvanometer. Dotted boxes enclose the completed instruments.

See Figs. 4.21 and 4.22 in your NCERT textbook

Worked example 6. A 3.00 V supply with no additional internal resistance drives a circuit containing a 3.00 Ω resistor and a current meter in series. Compare the current for a 60.00 Ω galvanometer alone, that galvanometer shunted by 0.02 Ω, and an ideal zero-resistance ammeter. The symbol Ω denotes ohm, the resistance unit.

Answer: With the galvanometer alone, total resistance is 63 Ω, giving 3/63 = 0.048 A. With the shunt, meter resistance is (60 × 0.02)/(60 + 0.02) ≈ 0.02 Ω, giving 3/3.02 = 0.99 A. The ideal ammeter gives 3/3 = 1.00 A.

Glossary

• Magnetic field — A vector field describing magnetic influence, measured through the force on a moving charge.
• Lorentz force — The total force exerted on a charged particle by electric and magnetic fields together.
• Current element — A very short directed segment of a conductor carrying current, used when calculating magnetic fields.
• Superposition — The principle that the resultant field equals the vector sum of the individual source fields.
• Amperian loop — A closed path chosen for applying the circuital relation between magnetic field and enclosed current.
• Solenoid — A conducting wire wound in closely spaced helical turns, producing an axial magnetic field inside.
• Centripetal force — The inward force towards the centre required to maintain a particle's circular motion.
• Pitch — The distance a particle travels along the magnetic field during one complete turn of its helix.
• Magnetic moment — A vector describing a current loop's magnetic strength and orientation, directed by the right-hand rule.
• Radial magnetic field — A field arrangement keeping the magnetic field in the galvanometer coil's plane as the coil turns.
• Current sensitivity — The angular deflection produced per unit current passing through the galvanometer coil.
• Shunt — A small parallel resistance that diverts most of the measured current away from a galvanometer coil.

Common errors and misconceptions

• Misconception: Every moving charge experiences a magnetic force. Correct: The force is zero when velocity is parallel or antiparallel to the magnetic field, even though the charge is moving.
• Misconception: A magnetic field increases a particle's speed by bending its path. Correct: Magnetic force is perpendicular to velocity and does no work; it changes direction without changing speed.
• Misconception: The angle in the coil-torque formula is measured from the coil's plane. Correct: It is measured between the field and the normal to the coil's plane.
• Misconception: Currents in the same direction repel because like charges repel. Correct: Parallel currents attract through magnetic interaction; oppositely directed currents repel.
• Misconception: A real finite solenoid has no external field. Correct: Its exterior field is weak; taking it as zero belongs to the ideal very-long-solenoid approximation.
• Misconception: Doubling galvanometer current sensitivity necessarily doubles voltage sensitivity. Correct: Voltage sensitivity also depends on coil resistance, which is likely to increase when more turns are added.
• Misconception: A shunt is connected in series when converting a galvanometer to an ammeter. Correct: The shunt goes in parallel with the coil, while the complete ammeter goes in series with the measured circuit.

Exam-style questions with model answers

Q1. State two conditions under which a charged particle experiences zero magnetic force in a non-zero uniform magnetic field. [2 marks]

1. A stationary charged particle experiences zero magnetic force because its speed is zero.
2. A moving charged particle also experiences zero magnetic force when its velocity is parallel or antiparallel to the magnetic field.

Q2. A tightly wound circular coil has 100 turns, radius 10 cm and current 1 A. Calculate the magnetic field at its centre in vacuum. Use μ₀ = 4π × 10⁻⁷ T m A⁻¹ and π = 3.14. [3 marks]

1. Convert the coil radius to metres: R = 0.10 m. All closely wound turns are treated as having this same radius.
2. The central field is B = μ₀NI/(2R), where N is the number of turns and I is the current.
3. Substituting the supplied values gives B = (4 × 3.14 × 10⁻⁷ × 100 × 1)/(2 × 0.10) = 6.28 × 10⁻⁴ T.

Q3. Two long straight parallel wires in vacuum carry 8.0 A and 5.0 A in the same direction, with separation 4.0 cm. Calculate the force on a 10 cm length of either wire and state its nature. Use μ₀/(2π) = 2 × 10⁻⁷ T m A⁻¹. [4 marks]

1. The separation is d = 0.040 m and the selected length is L = 0.10 m, expressed in compatible SI units.
2. The magnetic force magnitude follows F = μ₀I₁I₂L/(2πd), where I₁ and I₂ are the two wire currents.
3. Substitution gives F = (2 × 10⁻⁷ × 8.0 × 5.0 × 0.10)/0.040 = 2.0 × 10⁻⁵ N.
4. The force is attractive because the currents have the same direction; each selected wire segment is pulled towards the other wire.

Q4. Derive the magnetic field on the axis of a single circular loop in vacuum, then obtain its central field. The loop has radius R and steady current I; the observation point is distance x along the axis from its centre. Let μ₀ denote vacuum permeability. [5 marks]

1. Every short element dℓ lies a distance √(R² + x²) from the observation point. The element and its displacement to that point are perpendicular.
2. The Biot-Savart law therefore gives the magnitude of its small field contribution as dB = μ₀I dℓ/[4π(R² + x²)].
3. Resolve the field into axial and transverse components. The transverse components from diametrically opposite elements cancel, while the axial components reinforce.
4. Multiply each contribution by R/√(R² + x²) for its axial component. Summing dℓ around the circumference gives 2πR and hence B = μ₀IR²/[2(R² + x²)³ᐟ²].
5. At the centre x = 0, giving B = μ₀I/(2R). Curling the right-hand fingers along current identifies the axial direction with the thumb.

Q5. Explain the construction and working of a moving coil galvanometer and derive its current sensitivity. Let N be coil turns, A its area, B the radial field magnitude, Cₜ the spring's restoring torque per unit angular twist, I the current and φ the deflection. [5 marks]

1. A many-turn coil rotates about a fixed axis between magnetic poles. A cylindrical soft iron core strengthens the field and helps make it radial.
2. A spring supplies restoring torque and a pointer indicates deflection. The radial field remains in the coil's plane, keeping its angle with the normal at 90°.
3. The magnetic torque is therefore NIAB. The opposing spring torque is Cₜφ, proportional to the angular displacement from the zero-current position.
4. At steady deflection the torques balance: Cₜφ = NIAB. Thus φ = NIAB/Cₜ, so deflection is proportional to current for the given instrument.
5. Current sensitivity is deflection per unit current: Sᵢ = φ/I = NAB/Cₜ. More turns, greater area or stronger field increase it if the other factors remain unchanged.

Q6. A 3.00 V supply with no additional internal resistance feeds a 3.00 Ω resistor and a current meter in series. Find the current using (a) a 60.00 Ω galvanometer, (b) the same galvanometer with a 0.02 Ω parallel shunt and (c) an ideal zero-resistance ammeter. Explain the improvement in (b). [4 marks]

1. With the galvanometer alone, the total series resistance is 3.00 + 60.00 = 63.00 Ω. The current is 3.00/63.00 ≈ 0.048 A.
2. The shunted meter has resistance (60.00 × 0.02)/(60.00 + 0.02) ≈ 0.02 Ω. The circuit current becomes 3.00/3.02 ≈ 0.99 A.
3. An ideal ammeter adds no resistance, so the circuit current is 3.00/3.00 = 1.00 A.
4. The small parallel shunt diverts most current around the coil and greatly reduces the meter's resistance, bringing the measured circuit closer to its undisturbed current.

Q7. A square coil of side 10 cm has 20 turns and carries 12 A in a uniform magnetic field of 0.80 T. Its normal makes 30° with the field. Calculate the torque magnitude, using sin30° = 0.5. [4 marks]

1. The side length is 0.10 m, so the area of one turn is A = (0.10)² = 0.010 m².
2. The torque expression is τ = NIAB sinθ, where θ is the angle between the field and the coil's normal.
3. The given angle is already measured from the normal, so substitute N = 20, I = 12 A, B = 0.80 T and sinθ = 0.5.
4. The torque magnitude is therefore τ = 20 × 12 × 0.010 × 0.80 × 0.5 = 0.96 N m.

Q8. A galvanometer's number of turns and coil resistance are both doubled while coil area, magnetic field and spring torsional constant remain unchanged. Explain the changes in its current sensitivity and voltage sensitivity. [3 marks]

1. Current sensitivity is Sᵢ = NAB/Cₜ, where N is turns, A is area, B is field and Cₜ is the torsional constant. Doubling N therefore doubles Sᵢ.
2. Voltage sensitivity is Sᵥ = NAB/(CₜRɢ), where Rɢ is coil resistance. Both numerator and denominator double in this stated case.
3. Voltage sensitivity consequently remains unchanged. A rise in current sensitivity does not by itself guarantee a rise in voltage sensitivity because resistance also enters the voltage response.

Key takeaways

• Moving charges produce magnetic fields, and the resultant field of several sources is their vector sum.
• Magnetic force acts perpendicular to particle velocity, doing no work and leaving speed unchanged when acting alone.
• Biot-Savart adds contributions from current elements; Ampere's circuital law simplifies field calculations when the arrangement has suitable symmetry.
• The circular coil's field depends on current, turn count, radius and axial distance; specify the observation point before choosing the formula.
• Long parallel wires attract when their currents flow in the same direction and repel when the currents flow oppositely.
• A current loop in a uniform field has zero resultant force but can experience torque that turns its magnetic moment towards the field.
• A radial field makes galvanometer magnetic torque independent of coil deflection; the spring's opposing torque establishes the current reading.
• An ammeter uses a small parallel shunt, whereas a voltmeter uses a large series resistance; connect each complete instrument appropriately.

Test yourself

Why does reversing current reverse the compass orientation around a straight wire?

Reversing current reverses the magnetic field direction. The nearby compass needle consequently aligns in the opposite direction.

For a positive charge, what determines the direction of magnetic force?

The right-hand cross-product rule applied to velocity and magnetic field gives the force direction, perpendicular to both vectors.

What turns circular particle motion into a helical path in a uniform field?

A non-zero velocity component parallel to the field adds uniform motion along the field to circular motion perpendicular to it.

Why do transverse field components cancel on a circular loop's axis?

Diametrically opposite current elements produce equal and opposite transverse components at an axial point, while their axial components reinforce.

What condition allows the exterior solenoid field to be treated as zero?

It is the ideal very-long-solenoid approximation. For a finite solenoid the exterior field is weak rather than identically zero.

When is a current loop in stable equilibrium in a uniform field?

Its magnetic moment is parallel to the external field. A small angular displacement then produces a restoring torque.

Why does a galvanometer need a shunt to measure larger currents?

The small parallel resistance carries most of the current, limiting coil current while lowering the resistance of the complete ammeter.

Why is a large resistance added in series when making a voltmeter?

It reduces the current drawn by the instrument, limiting the disturbance of the potential difference being measured.