Question 1
A 100-turn circular coil of radius 8.0 cm carries 0.40 A. Find the magnetic field at its centre.
Steps only
B=μ₀NI/(2R)=(4π×10⁻⁷)(100)(0.40)/0.16=3.14×10⁻⁴ T
Answer
3.14×10⁻⁴ T
CLASS 12 · PHYSICS · CHAPTER 4
Moving charges and magnetism
Simple worked answers to all 13 exercises on magnetic fields, forces and coils.
13 worked answersComplete chapter answers
13 worked answers · short steps first, explanations when you need them
Chapter 4 Exercises
Textbook pages 135–136
Question 2
A long straight wire carries 35 A. Find the magnetic field 20 cm from it.
Steps only
B=μ₀I/(2πr)=(4π×10⁻⁷)(35)/(2π×0.20)=3.5×10⁻⁵ T
Answer
3.5×10⁻⁵ T
Question 3
A horizontal wire carries 50 A from north to south. Find the magnetic field 2.5 m east of the wire, including its direction.
Steps only
B=μ₀I/(2πr)=4.0×10⁻⁶ T; right-hand rule gives upward at the east side.
Answer
4.0×10⁻⁶ T upward
Question 4
An overhead wire carries 90 A from east to west. Find the field 1.5 m below it, including its direction.
Steps only
B=μ₀I/(2πr)=1.2×10⁻⁵ T; direction below a westward current is south.
Answer
1.2×10⁻⁵ T south
Question 5
A wire carrying 8 A makes a 30° angle with a 0.15 T magnetic field. Find the magnetic force per unit length.
Steps only
F/L=IB sinθ=8(0.15)(0.5)=0.60 N/m
Answer
0.60 N/m
Question 6
A 3.0 cm wire carries 10 A perpendicular to a 0.27 T magnetic field inside a solenoid. Find the force on the wire.
Steps only
F=ILB=10(0.030)(0.27)=0.081 N
Answer
0.081 N
Question 7
Parallel wires A and B carry 8.0 A and 5.0 A in the same direction and are 4.0 cm apart. Find the force on a 10 cm length of wire A.
Steps only
F=μ₀I₁I₂L/(2πd)=2.0×10⁻⁵ N; currents in the same direction attract.
Answer
2.0×10⁻⁵ N toward B
Question 8
An 80 cm solenoid has 5 layers of 400 turns each, diameter 1.8 cm, and carries 8.0 A. Estimate the field near its centre.
Steps only
n=2000/0.80=2500 m⁻¹; B=μ₀nI=2.51×10⁻² T
Answer
2.5×10⁻² T
Question 9
A 20-turn square coil of side 10 cm carries 12 A. Its normal makes 30° with a 0.80 T field. Find the torque.
Steps only
τ=NIAB sin30°=20(12)(0.010)(0.80)(0.5)=0.96 N m
Answer
0.96 N m
Question 10
Two moving-coil meters have: M₁: R₁=10 Ω, N₁=30, A₁=3.6×10⁻³ m², B₁=0.25 T; M₂: R₂=14 Ω, N₂=42, A₂=1.8×10⁻³ m², B₂=0.50 T. Their springs are identical. Find the ratios of (a) current sensitivity and (b) voltage sensitivity of M₂ to M₁.
Steps only
Current: (NAB)₂/(NAB)₁=1.4; voltage: 1.4(R₁/R₂)=1.0
Answer
1.4; 1.0
Question 11
An electron enters a 6.5 G uniform field at 4.8×10⁶ m/s, perpendicular to the field. Explain why its path is circular and find the orbit radius. Use e=1.5×10⁻¹⁹ C and mₑ=9.1×10⁻³¹ kg.
Steps only
r=mv/(eB)=(9.1×10⁻³¹)(4.8×10⁶)/[(1.5×10⁻¹⁹)(6.5×10⁻⁴)]=4.48 cm
Answer
4.48 cm; circular path
Question 12
For Exercise 4.11, find the electron's frequency of revolution. Does it depend on its speed? Explain.
Steps only
f=eB/(2πm)≈1.71×10⁷ Hz; speed is not in the formula.
Answer
17.1 MHz; independent of speed
Question 13
(a) A 30-turn coil of radius 8.0 cm carries 6.0 A in a 1.0 T field. The field makes 60° with the coil's normal. Find the counter-torque needed to keep it still. (b) Would the answer change for an irregular planar coil enclosing the same area?
Steps only
τ=NIπR²B sin60°≈3.14 N m; same enclosed area gives same torque.
Answer
3.14 N m; no change for same area