Chapter 4: Moving Charges and Magnetism
Key Formulas:
• Biot-Savart Law: dB = (μ₀/4π)(Idl × r/r³)
• Ampere's Law: ∮ B · dl = μ₀Ienc
• Force on charge: F = qvB sinθ
• Cyclotron radius: r = mv/(qB)
• Cyclotron frequency: f = qB/(2πm)
Q1. An electron moves with a speed of 3 × 107 m/s at an angle of 30° with the direction of a magnetic field of 0.1 T. Find the force on the electron.
Solution
F = qvB sinθ
= (1.6 × 10−19)(3 × 107)(0.1) sin 30°
= (1.6 × 10−19)(3 × 107)(0.1)(0.5)
= 2.4 × 10−13 N
Q2. Find the magnetic field at the centre of a circular coil of radius 10 cm carrying current 2A, with 50 turns.
Solution
B = μ₀NI/(2r)
= (4π × 10−7)(50)(2)/(2 × 0.1)
= 4π × 10−7 × 500
= 2π × 10−4 T ≈ 6.28 × 10−4 T
Q3. A proton moves in a circle of radius 10 cm with a speed of 106 m/s. Find the magnetic field. (m = 1.67 × 10−27 kg, q = 1.6 × 10−19 C)
Solution
r = mv/(qB)
B = mv/(qr) = (1.67 × 10−27)(106)/((1.6 × 10−19)(0.1))
= 1.67 × 10−21/(1.6 × 10−20)
= 0.104 T
Q4. A circular loop of radius R carries a current I. Show that at a point on its axis at distance x from the centre, B = μ₀IR²/2(R² + x²)3/2.
Solution
Using Biot-Savart law:
dB = (μ₀I dl sin 90°)/(4π)(R² + x²)
Only axial component survives:
dBaxial = dB · R/√(R² + x²)
B = ∮ dBaxial = (μ₀I)/(4π(R²+x²)) · R/√(R²+x²) · 2πR
= μ₀IR²/2(R² + x²)3/2
Q5. A charged particle of mass m and charge q is released from rest in a uniform electric field E and uniform magnetic field B (perpendicular to each other). Describe the path of the particle.
Solution
When E and B are perpendicular, the particle follows a cycloid path. Starting from rest, the electric field accelerates the particle, while the magnetic field deflects it. The path is a combination of circular motion superimposed on a linear drift, producing a cycloid.
Q6. An electron enters a region of uniform magnetic field B = 0.1 T with velocity v = 2 × 107 m/s perpendicular to the field. Find the time period of its circular motion.
Solution
T = 2πm/(qB)
= 2π(9.1 × 10−31)/((1.6 × 10−19)(0.1))
= 2π(9.1 × 10−31)/(1.6 × 10−20)
= 3.57 × 10−10 s ≈ 0.36 ns
Q7. A straight wire of length L carries a current I. Find the magnetic field at a perpendicular distance d from the midpoint of the wire.
Solution
Using Biot-Savart law and integrating:
B = (μ₀I)/(4πd)(sinθ₁ + sinθ₂)
For a long wire (L >> d), sinθ₁ ≈ sinθ₂ ≈ 1:
B = μ₀I/(2πd)
Q8. A solenoid of length 1 m has 1000 turns and carries a current of 5A. Find (i) the magnetic field inside (ii) the magnetic field at one end.
Solution
n = N/l = 1000/1 = 1000 turns/m
(i) Inside: B = μ₀nI = (4π × 10−7)(1000)(5)
= 20π × 10−4 = 6.28 × 10−3 T
(ii) At one end: B = μ₀nI/2 = 3.14 × 10−3 T
Q9. A charged particle of mass 10−26 kg and charge 10−16 C moves in a uniform magnetic field of 0.1 T with velocity 106 m/s perpendicular to the field. Find the radius of the circular path and the time period.
Solution
r = mv/(qB) = (10−26)(106)/((10−16)(0.1))
= 10−20/10−17 = 10−3 m = 1 mm
T = 2πm/(qB) = 2π(10−26)/((10−16)(0.1))
= 2π × 10−9 = 6.28 × 10−9 s = 6.28 ns
Q10. Explain the working principle of a cyclotron. What is the maximum kinetic energy of the ions accelerated in it?
Solution
A cyclotron accelerates charged particles using a combination of a uniform magnetic field and an alternating electric field across two D-shaped electrodes (Dees).
Working: A positive ion between the Dees is accelerated by the electric field. Inside each Dee, the magnetic field causes it to move in a semicircular path. The polarity of the Dees alternates to ensure the ion is accelerated each time it crosses the gap.
Maximum KE: Kmax = q²B²R²/(2m)
where R = radius of the Dees.
Q11. A galvanometer of resistance 100 Ω requires 1 mA for full-scale deflection. Convert it into an ammeter of range 0-1 A.
Solution
Shunt resistance: S = I₁R₁/(I − I₁)
= (0.001)(100)/(1 − 0.001)
= 0.1/0.999
= 0.1001 Ω
A low resistance of 0.1001 Ω is connected in parallel with the galvanometer.
Q12. A straight conductor of length L carries current I. Find the force on it when placed in a uniform magnetic field B at an angle θ with the field.
Solution
Force on a current-carrying conductor:
F = BIL sinθ
Direction: Given by Fleming's left-hand rule (or right-hand rule for cross product).
Q13. An electron enters a region of crossed electric and magnetic fields (E perpendicular to B). If the electron passes undeflected, find the velocity of the electron.
Solution
For undeflected motion: FE = FB
qE = qvB
v = E/B
This arrangement is called a velocity selector.
Q14. A circular coil of radius 10 cm has 200 turns and carries a current of 5A. Find the magnetic dipole moment of the coil.
Solution
M = NIA = NI(πr²)
= 200 × 5 × π(0.1)²
= 1000 × π × 0.01
= 10π A·m² ≈ 31.4 A·m²