Moving Charges and Magnetism

1 exercises10 questions solved

Exercise 4.1Magnetic Force, Biot-Savart Law and Ampere'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?

Solution

N = 100 turns, r = 8.0 cm = 0.08 m, I = 0.40 A Magnetic field at the centre of a circular coil: B = μ₀NI/(2r) = (4π × 10⁻⁷ × 100 × 0.40) / (2 × 0.08) = (4π × 10⁻⁷ × 40) / 0.16 = (160π × 10⁻⁷) / 0.16 = 1000π × 10⁻⁷ = π × 10⁻⁴ = 3.14 × 10⁻⁴ T B ≈ 3.14 × 10⁻⁴ T

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?

Solution

I = 35 A, r = 20 cm = 0.20 m Magnetic field due to a long straight wire: B = μ₀I/(2πr) = (4π × 10⁻⁷ × 35) / (2π × 0.20) = (4π × 35 × 10⁻⁷) / (2π × 0.20) = (4 × 35 × 10⁻⁷) / (2 × 0.20) = (140 × 10⁻⁷) / 0.40 = 350 × 10⁻⁷ = 3.5 × 10⁻⁵ T B = 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.

Solution

I = 50 A (north to south, i.e., in –y direction), r = 2.5 m B = μ₀I/(2πr) = (4π × 10⁻⁷ × 50) / (2π × 2.5) = (4 × 50 × 10⁻⁷) / (2 × 2.5) = (200 × 10⁻⁷) / 5 = 40 × 10⁻⁷ = 4 × 10⁻⁶ T Direction: Using right-hand thumb rule: current is toward south (–ŷ), point is to the east (+x̂). B = (μ₀I/2πr)(Î_current × r̂) Ĉurrent = –ŷ, r̂ = +x̂ –ŷ × x̂ = –(ŷ × x̂) = –(–ẑ) = +ẑ (vertically upward) B = 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?

Solution

I = 90 A (east to west, –x̂ direction), r = 1.5 m below the wire B = μ₀I/(2πr) = (4π × 10⁻⁷ × 90) / (2π × 1.5) = (4 × 90 × 10⁻⁷) / (2 × 1.5) = (360 × 10⁻⁷) / 3 = 120 × 10⁻⁷ = 1.2 × 10⁻⁵ T Direction: Current is west (–x̂), point is below (–ẑ direction from wire) Using right-hand rule: (–x̂) × (–ẑ) = x̂ × ẑ = –ŷ (i.e., toward south) B = 1.2 × 10⁻⁵ T directed toward 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?

Solution

I = 8 A, B = 0.15 T, θ = 30° Force per unit length on a current-carrying wire: F/l = BIsinθ = 0.15 × 8 × sin30° = 0.15 × 8 × 0.5 = 0.6 N/m Force per unit length = 0.6 N/m

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?

Solution

l = 3.0 cm = 0.03 m, I = 10 A, B = 0.27 T The wire is perpendicular to the field, so θ = 90° F = BIl sinθ = BIl (since θ = 90°) = 0.27 × 10 × 0.03 = 8.1 × 10⁻² N = 0.081 N Force on the wire = 8.1 × 10⁻² N

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.

Solution

I_A = 8 A, I_B = 5 A, d = 4.0 cm = 0.04 m, l = 10 cm = 0.1 m Force per unit length between two parallel wires: F/l = μ₀I_AI_B/(2πd) = (4π × 10⁻⁷ × 8 × 5) / (2π × 0.04) = (4 × 10⁻⁷ × 8 × 5) / (2 × 0.04) = (160 × 10⁻⁷) / 0.08 = 2000 × 10⁻⁷ = 2 × 10⁻⁴ N/m Force on 10 cm (0.1 m) of wire A: F = (F/l) × l = 2 × 10⁻⁴ × 0.1 = 2 × 10⁻⁵ N The force is attractive (currents in same direction). F = 2 × 10⁻⁵ N (attractive)

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.

Solution

Length L = 80 cm = 0.8 m, layers = 5, turns per layer = 400 Total turns N = 5 × 400 = 2000 I = 8.0 A Number of turns per unit length: n = N/L = 2000/0.8 = 2500 turns/m Magnetic field inside solenoid: B = μ₀nI = 4π × 10⁻⁷ × 2500 × 8.0 = 4π × 10⁻⁷ × 20000 = 4π × 2 × 10⁻³ = 8π × 10⁻³ = 25.12 × 10⁻³ ≈ 2.5 × 10⁻² T B ≈ 2.5 × 10⁻² T

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?

Solution

Side a = 10 cm = 0.1 m, N = 20 turns, I = 12 A, θ = 30°, B = 0.80 T Area of coil: A = a² = (0.1)² = 0.01 m² Magnetic moment: M = NIA = 20 × 12 × 0.01 = 2.4 A·m² Torque: τ = MBsinθ = 2.4 × 0.80 × sin30° = 2.4 × 0.80 × 0.5 = 0.96 N·m The torque experienced by the coil is 0.96 N·m.

Q10

A galvanometer coil has a resistance of 12 Ω and the metre shows full scale deflection for a current of 3 mA. How will you convert the metre into a voltmeter of range 0 to 18 V, and an ammeter of range 0 to 6 A? What are the values of the resistors you would use?

Solution

Galvanometer: G = 12 Ω, I_g = 3 mA = 3 × 10⁻³ A Conversion to Voltmeter (range 0–18 V): Connect a high resistance R in series with the galvanometer. At full scale: V = I_g(G + R) 18 = 3 × 10⁻³ × (12 + R) 6000 = 12 + R R = 5988 Ω ≈ 5988 Ω Conversion to Ammeter (range 0–6 A): Connect a small shunt resistance S in parallel. At full scale: I_g × G = (I – I_g) × S 3 × 10⁻³ × 12 = (6 – 3 × 10⁻³) × S 0.036 = 5.997 × S S = 0.036/5.997 ≈ 6 × 10⁻³ Ω = 0.006 Ω Voltmeter: series resistance = 5988 Ω Ammeter: shunt resistance ≈ 0.006 Ω

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Moving Charges and Magnetism

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