Exercises
1. 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?
Answer
Given: N = 100, R = 0.08 m, I = 0.40 A.
B = μ₀NI/(2R)
⇒ B = (4π × 10⁻⁷ × 100 × 0.40)/(2 × 0.08)
⇒ B = (5.03 × 10⁻⁵)/0.16
⇒ B ≈ 3.1 × 10⁻⁴ T
2. 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?
Answer
B = μ₀I/(2πr)
⇒ B = (2 × 10⁻⁷ × 35)/0.20
⇒ B = 3.5 × 10⁻⁵ T
3. 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.
Answer
B = μ₀I/(2πr)
⇒ B = (2 × 10⁻⁷ × 50)/2.5
⇒ B = 4 × 10⁻⁶ T
Direction (right-hand thumb rule: thumb pointing south, along the current): at a point to the east the field lines point vertically upward.
4. 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?
Answer
B = μ₀I/(2πr)
⇒ B = (2 × 10⁻⁷ × 90)/1.5
⇒ B = 1.2 × 10⁻⁵ T
Direction (right-hand thumb rule, thumb pointing west): at a point directly below the line the field is horizontal, towards the south.
5. What is the magnitude of magnetic force per unit length on a wire carrying a current of 8 A and making an angle of 30º with the direction of a uniform magnetic field of 0.15 T?
Answer
F/l = IB sin θ
⇒ F/l = 8 × 0.15 × sin 30°
⇒ F/l = 1.2 × 0.5
⇒ F/l = 0.6 N m⁻¹
6. 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?
Answer
The field inside the solenoid is along its axis, so θ = 90°.
F = IlB sin θ
⇒ F = 10 × 0.03 × 0.27 × 1
⇒ F = 8.1 × 10⁻² N
The force is perpendicular to both the wire and the solenoid axis; its sense follows Fleming's left-hand rule.
7. 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.
Answer
F = μ₀I_AI_B l/(2πd)
⇒ F = (2 × 10⁻⁷ × 8.0 × 5.0 × 0.10)/0.04
⇒ F = (8 × 10⁻⁷)/0.04
⇒ F = 2 × 10⁻⁵ N
Currents are in the same direction, so the force is attractive: wire A is pulled towards wire B (perpendicular to A).
8. 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.
Answer
Total turns N = 5 × 400 = 2000; length l = 0.80 m
⇒ n = N/l = 2500 turns per metre.
B = μ₀nI
⇒ B = 4π × 10⁻⁷ × 2500 × 8.0
⇒ B ≈ 2.5 × 10⁻² T
(The diameter does not matter, since the solenoid is long compared with its diameter.)
9. 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 30º with the direction of a uniform horizontal magnetic field of magnitude 0.80 T. What is the magnitude of torque experienced by the coil?
Answer
A = (0.10)² = 10⁻² m², θ = 30° (between normal and field).
τ = NIAB sin θ
⇒ τ = 20 × 12 × 10⁻² × 0.80 × 0.5
⇒ τ = 0.96 N m
10. 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.
Answer
Current sensitivity I_s = NAB/k; voltage sensitivity V_s = NAB/(kR).
(a) I_s2/I_s1 = (N₂A₂B₂)/(N₁A₁B₁)
⇒ (42 × 1.8 × 10⁻³ × 0.50)/(30 × 3.6 × 10⁻³ × 0.25)
⇒ 0.0378/0.027
⇒ 1.4
(b) V_s2/V_s1 = (N₂A₂B₂R₁)/(N₁A₁B₁R₂)
⇒ 1.4 × (10/14)
⇒ 1
11. 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)
Answer
The magnetic force F = –e(v × B) is always perpendicular to the velocity. It therefore changes only the direction of motion, not the speed, and has constant magnitude evB. A constant-magnitude force always perpendicular to the velocity provides centripetal force, so the electron moves in a circle.
evB = mv²/r
⇒ r = mv/(eB)
⇒ r = (9.1 × 10⁻³¹ × 4.8 × 10⁶)/(1.5 × 10⁻¹⁹ × 6.5 × 10⁻⁴)
⇒ r = (4.37 × 10⁻²⁴)/(9.75 × 10⁻²³)
⇒ r ≈ 4.5 × 10⁻² m = 4.5 cm
(Note: the book prints e = 1.5 × 10⁻¹⁹ C; with the standard value e = 1.6 × 10⁻¹⁹ C the radius is r ≈ 4.2 cm.)
12. 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.
Answer
ν = eB/(2πm)
⇒ ν = (1.5 × 10⁻¹⁹ × 6.5 × 10⁻⁴)/(2 × 3.14 × 9.1 × 10⁻³¹)
⇒ ν = (9.75 × 10⁻²³)/(5.72 × 10⁻³⁰)
⇒ ν ≈ 1.7 × 10⁷ Hz ≈ 17 MHz (≈ 18 MHz with e = 1.6 × 10⁻¹⁹ C)
No, it does not depend on speed. Since r = mv/(eB), a faster electron moves on a proportionally larger circle; the time period T = 2πr/v = 2πm/(eB) stays the same (valid for non-relativistic speeds).
13. (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.)
Answer
(a) A = πr² = 3.14 × (0.08)² = 2.01 × 10⁻² m².
τ = NIAB sin θ
⇒ τ = 30 × 6.0 × 2.01 × 10⁻² × 1.0 × sin 60°
⇒ τ = 3.62 × 0.866
⇒ τ ≈ 3.13 N m
An equal and opposite counter torque of about 3.1 N m must be applied.
(b) No. The torque τ = NIAB sin θ depends only on the area of the planar coil, not on its shape, so the answer is unchanged.
Important Extra Questions
Write the expression for the Lorentz force on a charge q moving in electric and magnetic fields.
Answer
F = q[E + (v × B)]. The magnetic part q(v × B) is perpendicular to the velocity and does no work on the charge.
State the Biot–Savart law.
Answer
The field due to a current element I dl at a point at position r is dB = (μ₀/4π) I(dl × r)/r³, i.e., magnitude dB = (μ₀/4π) I dl sin θ/r².
State Ampere's circuital law.
Answer
The line integral of B around any closed loop equals μ₀ times the total current threading the loop: ∮B·dl = μ₀I.
How is a galvanometer converted into an ammeter and into a voltmeter?
Answer
Ammeter: connect a small resistance (shunt) r_s in parallel with the galvanometer. Voltmeter: connect a large resistance R in series with it, R = V/I_g – G.
Define one ampere in terms of the force between parallel currents.
Answer
One ampere is the steady current which, flowing in each of two very 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 between them.