RowQ
The Vault
RowQ
The Vault
CBSE Class 12 Physics · 11 questions · 27 marks
A current is nothing but charge in motion, and charge in motion wraps the space around it in a magnetic field. This chapter hands you two tools for finding that field — the Biot–Savart law for awkward shapes and Ampere's circuital law for symmetric ones — then flips the question to ask what force the field pushes back with. Cyclotrons, solenoids and the moving coil galvanometer all turn out to be the same three formulas in different costumes.
A proton and an alpha particle, having the same momentum, enter a uniform magnetic field at right angles to the field. The ratio of the radii of their circular paths (proton : alpha) is:
Answer
2 : 1 — the radius is r = mv/qB = p/qB. Since both particles carry the same momentum p and move in the same field B, r depends only on the charge, r ∝ 1/q. The alpha particle has charge 2e against the proton's e, so r_p/r_α = 2/1. Mass does not enter once momentum is fixed.
A flat circular coil of 50 turns and radius 10 cm carries a current of 2.0 A. The magnetic field at its centre is nearest to:
Answer
6.3×10⁻⁴ T — using B = μ₀NI/2R, B = (4π×10⁻⁷ × 50 × 2.0)/(2 × 0.10) = (4π×10⁻⁷ × 100)/0.20 = 1.2566×10⁻⁴/0.20 = 6.28×10⁻⁴ T. The field is directed along the axis, its sense given by curling the right hand fingers along the current.
An electron is projected into a uniform magnetic field with its velocity exactly along the direction of the field. The electron will:
Answer
Continue in a straight line with unchanged speed — the magnetic force is F = qvB sinθ, and here θ = 0°, so sinθ = 0 and the force vanishes. With no force acting, the electron obeys Newton's first law and travels straight on at constant speed. A helical path would require a velocity component across the field.
A solenoid 40 cm long is wound with 800 turns and carries a steady current of 2.5 A. The magnetic field near the middle of its axis is closest to:
Answer
6.3×10⁻³ T — the number of turns per metre is n = 800/0.40 = 2000 m⁻¹. Then B = μ₀nI = 4π×10⁻⁷ × 2000 × 2.5 = 4π×10⁻⁷ × 5000 = 6.28×10⁻³ T. The field inside a long solenoid is uniform and depends on turns per unit length, not the total number of turns.
Assertion (A): The time period of a charged particle moving in a circular path in a uniform magnetic field is independent of its speed. Reason (R): The magnetic force on a moving charge always acts perpendicular to its velocity, so it changes the direction of motion but never the speed.
Answer
Both A and R are true but R is not the correct explanation of A — R correctly states why the speed stays constant, but the independence of the period comes from a different step. Since qvB supplies the centripetal force, r = mv/qB, so T = 2πr/v = 2πm/qB, and the v cancels. A faster particle simply sweeps a proportionally larger circle in the same time.
State the Biot–Savart law and use it to obtain an expression for the magnetic field at the centre of a circular coil of N turns, radius R, carrying a current I.
Answer
Biot–Savart law: the magnetic field dB produced at a point by a small current element I dl is directly proportional to the current, to the length of the element and to sinθ (θ being the angle between dl and the line joining the element to the point), and inversely proportional to the square of the distance r: dB = (μ₀/4π)·(I dl sinθ)/r², directed along dl × r. Application to a circular coil: Take the point at the centre O of a single turn of radius R. Every current element dl on the ring is perpendicular to the radius drawn to O, so θ = 90° and sinθ = 1, and every element is the same distance R away. dB = (μ₀/4π)·I dl/R². By the right hand rule, the contribution of every element points the same way along the axis, so the magnitudes simply add: B = ∮(μ₀/4π)·I dl/R² = (μ₀I/4πR²)∮dl = (μ₀I/4πR²)(2πR) B = μ₀I/2R. For N closely wound turns the contributions add, giving B = μ₀NI/2R.
Two long straight parallel wires, 8.0 cm apart, carry currents of 6.0 A and 10.0 A in the same direction. Calculate the force per unit length between them and state whether it is attractive or repulsive.
Answer
The force per unit length between long parallel currents is F/l = μ₀I₁I₂/2πd = 2×10⁻⁷ × I₁I₂/d. Here I₁ = 6.0 A, I₂ = 10.0 A, d = 8.0 cm = 0.080 m. F/l = 2×10⁻⁷ × (6.0 × 10.0)/0.080 F/l = 2×10⁻⁷ × 60/0.080 F/l = 1.2×10⁻⁵/0.080 = 1.5×10⁻⁴ N/m. Since the currents flow in the same direction, the force is attractive; the wires are pulled towards each other.
A moving coil galvanometer of resistance 60 Ω gives a full scale deflection for a current of 4.0 mA. Explain how it can be converted into an ammeter reading up to 3.0 A, and calculate the required resistance.
Answer
A galvanometer is converted into an ammeter by joining a small resistance, called a shunt, in parallel with it. The shunt carries most of the current so that the delicate coil never receives more than its full scale current, and the low combined resistance means the meter barely disturbs the circuit it is placed in. At full scale the potential difference across the galvanometer equals that across the shunt: I_g G = (I - I_g) S S = I_g G/(I - I_g) Here I_g = 4.0 mA = 4.0×10⁻³ A, G = 60 Ω, I = 3.0 A. S = (4.0×10⁻³ × 60)/(3.0 - 4.0×10⁻³) S = 0.24/2.996 S = 0.080 Ω. So a shunt of about 0.080 Ω connected in parallel converts it into a 0 to 3 A ammeter.
Using the Biot–Savart law, derive an expression for the magnetic field at a point on the axis of a circular current loop of radius R carrying current I, at a distance x from the centre. Hence deduce the field at the centre. Calculate the axial field 4.0 cm from the centre of a 20-turn coil of radius 3.0 cm carrying 1.5 A.
Answer
Derivation: Let the loop of radius R lie in the y-z plane with its centre O at the origin, carrying current I. Take the point P on the x-axis at distance x from O. Consider a small current element I dl on the loop. Its distance from P is r = √(R² + x²), and dl is perpendicular to r, so θ = 90° and sinθ = 1. By the Biot–Savart law the magnitude of its contribution is dB = (μ₀/4π)·I dl/(R² + x²). The direction of dB is perpendicular to the plane containing dl and r, so dB has a component along the axis, dB cosα, and a component perpendicular to the axis, dB sinα, where cosα = R/√(R² + x²). Now take the diametrically opposite element. Its perpendicular component is equal and opposite, so all perpendicular components cancel in pairs around the loop, while all axial components add. Therefore B = ∮dB cosα = ∮(μ₀/4π)·I dl/(R² + x²) × R/√(R² + x²) B = (μ₀ I R/4π(R² + x²)^(3/2))∮dl With ∮dl = 2πR: B = μ₀IR²/2(R² + x²)^(3/2), directed along the axis. For N turns, B = μ₀NIR²/2(R² + x²)^(3/2). At the centre: putting x = 0 gives B = μ₀IR²/2R³ = μ₀I/2R, or μ₀NI/2R for N turns, which agrees with the earlier result. Numerical: N = 20, I = 1.5 A, R = 3.0×10⁻² m, x = 4.0×10⁻² m. R² + x² = (9.0×10⁻⁴) + (16.0×10⁻⁴) = 25.0×10⁻⁴ m². √(R² + x²) = 5.0×10⁻² m, so (R² + x²)^(3/2) = (5.0×10⁻²)³ = 1.25×10⁻⁴ m³. Numerator: μ₀NIR² = 4π×10⁻⁷ × 20 × 1.5 × 9.0×10⁻⁴ = 4π×10⁻⁷ × 2.7×10⁻² = 3.393×10⁻⁸. B = 3.393×10⁻⁸/(2 × 1.25×10⁻⁴) = 3.393×10⁻⁸/2.5×10⁻⁴ B = 1.36×10⁻⁴ T, directed along the axis of the coil.
(a) State Ampere's circuital law and use it to derive an expression for the magnetic field inside a long current-carrying solenoid. (b) Derive the expression for the force per unit length between two long parallel current-carrying wires and use it to define the ampere. (c) Find the magnetic moment of a 120-turn rectangular coil of sides 8.0 cm and 5.0 cm carrying 0.75 A, and the maximum torque it can feel in a field of 0.30 T.
Answer
(a) Ampere's circuital law: the line integral of the magnetic field around any closed loop equals μ₀ times the net current threading that loop: ∮B·dl = μ₀I(enclosed). Derivation for a solenoid: Consider a long solenoid with n turns per unit length carrying current I. The field inside is uniform and along the axis; outside a long solenoid it is negligibly weak. Choose a rectangular Amperian loop PQRS with side PQ of length L lying along the axis inside the solenoid, side RS of the same length outside, and the two short sides perpendicular to the axis. Along PQ: B is parallel to the path, contributing BL. Along RS: B is essentially zero outside, contributing 0. Along the two short sides: B is perpendicular to the path inside and zero outside, contributing 0. So ∮B·dl = BL. Number of turns enclosed by the loop = nL, so I(enclosed) = nLI. Applying the law: BL = μ₀nLI, giving B = μ₀nI. The field inside a long solenoid therefore depends only on the current and on the turns per unit length, not on the radius or on the position within the interior. (b) Two long straight wires a distance d apart carry currents I₁ and I₂ in the same direction. Wire 1 sets up at the location of wire 2 a field B₁ = μ₀I₁/2πd, directed perpendicular to wire 2. The force on a length l of wire 2 is F = B₁I₂l = μ₀I₁I₂l/2πd, so the force per unit length is F/l = μ₀I₁I₂/2πd. Applying Fleming's left hand rule shows the force is attractive for parallel currents and repulsive for antiparallel currents. Definition of the ampere: one ampere is that steady current which, flowing in each of two infinitely long straight parallel conductors of negligible cross-section placed 1 m apart in vacuum, produces a force of 2×10⁻⁷ newton per metre of length on each conductor. (c) Area A = 0.080 × 0.050 = 4.0×10⁻³ m². Magnetic moment m = NIA = 120 × 0.75 × 4.0×10⁻³ = 0.36 A m². Maximum torque occurs at θ = 90°: τ(max) = mB = 0.36 × 0.30 = 0.108 N m ≈ 0.11 N m.
Read the passage and answer the questions that follow: A college physics club builds a simple mass analyser. Singly charged positive ions are produced, accelerated, and injected with a speed of 3.0×10⁵ m/s into a chamber where a uniform magnetic field of 0.25 T acts perpendicular to their velocity. The ions bend in semicircles and strike a detector plate placed alongside the entry slit, so the distance from the slit to the strike point equals the diameter of the ion's path. The club tests a sample containing two isotopes, of masses 3.30×10⁻²⁶ kg and 3.65×10⁻²⁶ kg. (Take charge on each ion = 1.6×10⁻¹⁹ C.) (a) Calculate the radius of the path of the lighter isotope. (b) Calculate the time taken by an ion of the lighter isotope to complete one full circle. (c) Find the separation between the points where the two isotopes strike the detector. (d) The club worries that the magnetic field will speed the ions up before they land. Explain why it cannot.
Answer
(a) The magnetic force supplies the centripetal force, qvB = mv²/r, so r = mv/qB. r₁ = (3.30×10⁻²⁶ × 3.0×10⁵)/(1.6×10⁻¹⁹ × 0.25) Numerator = 9.90×10⁻²¹; denominator = 4.0×10⁻²⁰. r₁ = 0.248 m ≈ 24.8 cm. (b) T = 2πm/qB = (2 × 3.14 × 3.30×10⁻²⁶)/(4.0×10⁻²⁰) Numerator = 2.073×10⁻²⁵. T = 5.2×10⁻⁶ s, that is about 5.2 μs. Note this does not depend on the speed of the ion. (c) For the heavier isotope, r₂ = (3.65×10⁻²⁶ × 3.0×10⁵)/(4.0×10⁻²⁰) = 1.095×10⁻²⁰/4.0×10⁻²⁰ = 0.274 m. Each ion lands at a distance equal to its diameter from the slit, so the separation is Δ = 2r₂ - 2r₁ = 2(0.274 - 0.248) = 2 × 0.026 = 0.052 m ≈ 5.2 cm. The heavier isotope lands 5.2 cm further from the slit, which is how the analyser separates them. (d) The magnetic force F = q(v × B) is always perpendicular to the velocity, so at every instant the force is at right angles to the displacement and the work done, W = F·d, is zero. With no work done, the kinetic energy and hence the speed of an ion cannot change; the field only turns the velocity vector, which is exactly what makes the path a circle of constant radius.
RowQ generates fresh questions on Moving Charges and Magnetism, marks your answers, and explains every step.
Start free