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Moving Charges and Magnetism NCERT Solutions, Complete PDF Content and Important Questions
Welcome to your complete guide for CBSE Class 12 Physics Chapter 4! In "Moving Charges and Magnetism," you will discover how electric currents produce magnetic fields and how moving charges behave within them. Mastering this chapter is highly essential for scoring top marks in your 2026 board exams and clearing competitive tests like JEE and NEET. Let's make physics simple, clear your doubts, and boost your exam preparation!
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ChapterMoving Charges and Magnetism
SubjectPhysics
Class12
BoardCBSE / State Boards
DifficultyModerate to High
Exam Weightage7-8 Marks (Combined)
Learning Objectives
After completing this chapter, students will be able to:
- Understand the concept of magnetic fields and how they are generated by moving charges.
- Apply the Biot-Savart Law and Ampere's Circuital Law to calculate magnetic fields for different current distributions.
- Calculate the magnetic force acting on a moving charge and a current-carrying conductor (Lorentz Force).
- Analyze the motion of charged particles in combined electric and magnetic fields.
- Understand the principle, construction, and working of a Moving Coil Galvanometer. Convert a galvanometer into an ammeter and a voltmeter.
Key Concepts, Definitions and Formulas
- Magnetic Field ($\mathbf{B}$): The space around a magnet or a current-carrying conductor where its magnetic influence can be felt.
- Lorentz Force: The total force experienced by a charge $q$ moving with velocity $\mathbf{v}$ in the presence of both electric field $\mathbf{E}$ and magnetic field $\mathbf{B}$.
$$\mathbf{F} = q(\mathbf{E} + \mathbf{v} \times \mathbf{B})$$ - Biot-Savart Law: Gives the magnetic field $d\mathbf{B}$ due to a small current element $I d\mathbf{l}$.
$$d\mathbf{B} = \frac{\mu_0}{4\pi} \frac{I d\mathbf{l} \times \mathbf{r}}{r^3}$$ - Magnetic Field at the center of a circular coil:
$$B = \frac{\mu_0 N I}{2R}$$ - Ampere's Circuital Law: The line integral of the magnetic field around any closed loop is equal to $\mu_0$ times the total current threading the loop.
$$\oint \mathbf{B} \cdot d\mathbf{l} = \mu_0 I$$ - Magnetic Field of a Solenoid: For a long solenoid with $n$ turns per unit length.
$$B = \mu_0 n I$$ - Force on a Current-Carrying Conductor:
$$\mathbf{F} = I(\mathbf{l} \times \mathbf{B})$$ - Force between Two Parallel Currents: Two parallel wires carrying currents in the same direction attract each other; in opposite directions, they repel. Force per unit length:
$$f = \frac{\mu_0 I_1 I_2}{2\pi d}$$ - Magnetic Dipole Moment ($\mathbf{m}$): For a current loop with area $\mathbf{A}$ and $N$ turns.
$$\mathbf{m} = N I \mathbf{A}$$
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Full NCERT Solutions
(Note: These are the standard rationalized exercise questions for CBSE Board Exam 2026)
NCERT Q1: 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:
Step 1: Identify given values.
Number of turns, $N = 100$
Radius of coil, $r = 8.0 \text{ cm} = 0.08 \text{ m}$
Current, $I = 0.40 \text{ A}$
Permeability of free space, $\mu_0 = 4\pi \times 10^{-7} \text{ T m/A}$
Step 2: Apply the formula for magnetic field at the center of a coil.
$$B = \frac{\mu_0 N I}{2r}$$
Step 3: Substitute the values and calculate.
$$B = \frac{(4\pi \times 10^{-7}) \times 100 \times 0.40}{2 \times 0.08}$$
$$B = \frac{4 \times 3.14 \times 10^{-7} \times 40}{0.16}$$
$$B = 3.14 \times 10^{-4} \text{ T}$$
Final Answer: The magnitude of the magnetic field is $3.14 \times 10^{-4} \text{ T}$.
Step 1: Identify given values.
Number of turns, $N = 100$
Radius of coil, $r = 8.0 \text{ cm} = 0.08 \text{ m}$
Current, $I = 0.40 \text{ A}$
Permeability of free space, $\mu_0 = 4\pi \times 10^{-7} \text{ T m/A}$
Step 2: Apply the formula for magnetic field at the center of a coil.
$$B = \frac{\mu_0 N I}{2r}$$
Step 3: Substitute the values and calculate.
$$B = \frac{(4\pi \times 10^{-7}) \times 100 \times 0.40}{2 \times 0.08}$$
$$B = \frac{4 \times 3.14 \times 10^{-7} \times 40}{0.16}$$
$$B = 3.14 \times 10^{-4} \text{ T}$$
Final Answer: The magnitude of the magnetic field is $3.14 \times 10^{-4} \text{ T}$.
NCERT Q2: 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:
Step 1: Identify given values.
Current, $I = 35 \text{ A}$
Distance from the wire, $r = 20 \text{ cm} = 0.2 \text{ m}$
Step 2: Apply the formula for an infinite straight wire.
$$B = \frac{\mu_0 I}{2\pi r}$$
Step 3: Calculate the field.
$$B = \frac{(4\pi \times 10^{-7}) \times 35}{2\pi \times 0.2}$$
$$B = \frac{2 \times 10^{-7} \times 35}{0.2}$$
$$B = 3.5 \times 10^{-5} \text{ T}$$
Final Answer: The magnetic field is $3.5 \times 10^{-5} \text{ T}$.
Step 1: Identify given values.
Current, $I = 35 \text{ A}$
Distance from the wire, $r = 20 \text{ cm} = 0.2 \text{ m}$
Step 2: Apply the formula for an infinite straight wire.
$$B = \frac{\mu_0 I}{2\pi r}$$
Step 3: Calculate the field.
$$B = \frac{(4\pi \times 10^{-7}) \times 35}{2\pi \times 0.2}$$
$$B = \frac{2 \times 10^{-7} \times 35}{0.2}$$
$$B = 3.5 \times 10^{-5} \text{ T}$$
Final Answer: The magnetic field is $3.5 \times 10^{-5} \text{ T}$.
NCERT Q3: 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:
Step 1: Note the given values.
$I = 50 \text{ A}$
$r = 2.5 \text{ m}$
Step 2: Calculate magnitude.
$$B = \frac{\mu_0 I}{2\pi r}$$
$$B = \frac{4\pi \times 10^{-7} \times 50}{2\pi \times 2.5}$$
$$B = \frac{2 \times 10^{-7} \times 50}{2.5} = 4 \times 10^{-6} \text{ T}$$
Step 3: Determine direction.
Using the Right-Hand Thumb Rule, point your thumb in the direction of current (North to South). Your fingers will curl upwards at a point to the East.
Final Answer: $4 \times 10^{-6} \text{ T}$ directed vertically upward.
Step 1: Note the given values.
$I = 50 \text{ A}$
$r = 2.5 \text{ m}$
Step 2: Calculate magnitude.
$$B = \frac{\mu_0 I}{2\pi r}$$
$$B = \frac{4\pi \times 10^{-7} \times 50}{2\pi \times 2.5}$$
$$B = \frac{2 \times 10^{-7} \times 50}{2.5} = 4 \times 10^{-6} \text{ T}$$
Step 3: Determine direction.
Using the Right-Hand Thumb Rule, point your thumb in the direction of current (North to South). Your fingers will curl upwards at a point to the East.
Final Answer: $4 \times 10^{-6} \text{ T}$ directed vertically upward.
NCERT Q4: 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:
Step 1: Note the given values.
$I = 90 \text{ A}$
$r = 1.5 \text{ m}$
Step 2: Calculate magnitude.
$$B = \frac{\mu_0 I}{2\pi r} = \frac{4\pi \times 10^{-7} \times 90}{2\pi \times 1.5} = 1.2 \times 10^{-5} \text{ T}$$
Step 3: Determine direction.
By Right-Hand Thumb Rule (thumb pointing East to West), below the wire, the fingers point towards the South.
Final Answer: $1.2 \times 10^{-5} \text{ T}$ towards the South.
Step 1: Note the given values.
$I = 90 \text{ A}$
$r = 1.5 \text{ m}$
Step 2: Calculate magnitude.
$$B = \frac{\mu_0 I}{2\pi r} = \frac{4\pi \times 10^{-7} \times 90}{2\pi \times 1.5} = 1.2 \times 10^{-5} \text{ T}$$
Step 3: Determine direction.
By Right-Hand Thumb Rule (thumb pointing East to West), below the wire, the fingers point towards the South.
Final Answer: $1.2 \times 10^{-5} \text{ T}$ towards the South.
NCERT Q5: What is the magnitude of magnetic force per unit length on a wire carrying a current of 8 A and making an angle of $30^\circ$ with the direction of a uniform magnetic field of 0.15 T?
Answer:
Step 1: Identify given values.
Current, $I = 8 \text{ A}$
Angle, $\theta = 30^\circ$
Magnetic field, $B = 0.15 \text{ T}$
Step 2: Apply Force per unit length formula.
$$f = \frac{F}{L} = I B \sin\theta$$
Step 3: Substitute the values and calculate.
$$f = 8 \times 0.15 \times \sin(30^\circ)$$
$$f = 1.2 \times 0.5 = 0.6 \text{ N/m}$$
Final Answer: The force per unit length is $0.6 \text{ N/m}$.
Step 1: Identify given values.
Current, $I = 8 \text{ A}$
Angle, $\theta = 30^\circ$
Magnetic field, $B = 0.15 \text{ T}$
Step 2: Apply Force per unit length formula.
$$f = \frac{F}{L} = I B \sin\theta$$
Step 3: Substitute the values and calculate.
$$f = 8 \times 0.15 \times \sin(30^\circ)$$
$$f = 1.2 \times 0.5 = 0.6 \text{ N/m}$$
Final Answer: The force per unit length is $0.6 \text{ N/m}$.
NCERT Q6: 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:
Step 1: Identify given values.
Length, $L = 3.0 \text{ cm} = 0.03 \text{ m}$
Current, $I = 10 \text{ A}$
Magnetic field, $B = 0.27 \text{ T}$
Angle, $\theta = 90^\circ$ (perpendicular to axis)
Step 2: Apply the force formula on the wire.
$$F = I L B \sin\theta$$
Step 3: Substitute the values and calculate.
$$F = 10 \times 0.03 \times 0.27 \times \sin(90^\circ)$$
$$F = 0.3 \times 0.27 \times 1 = 0.081 \text{ N}$$
Final Answer: The magnetic force is $8.1 \times 10^{-2} \text{ N}$.
Step 1: Identify given values.
Length, $L = 3.0 \text{ cm} = 0.03 \text{ m}$
Current, $I = 10 \text{ A}$
Magnetic field, $B = 0.27 \text{ T}$
Angle, $\theta = 90^\circ$ (perpendicular to axis)
Step 2: Apply the force formula on the wire.
$$F = I L B \sin\theta$$
Step 3: Substitute the values and calculate.
$$F = 10 \times 0.03 \times 0.27 \times \sin(90^\circ)$$
$$F = 0.3 \times 0.27 \times 1 = 0.081 \text{ N}$$
Final Answer: The magnetic force is $8.1 \times 10^{-2} \text{ N}$.
NCERT Q7: 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:
Step 1: Identify given values.
$I_A = 8.0 \text{ A}$, $I_B = 5.0 \text{ A}$
Distance, $d = 4.0 \text{ cm} = 0.04 \text{ m}$
Length of section, $L = 10 \text{ cm} = 0.1 \text{ m}$
Step 2: Apply the formula for force between parallel wires.
$$F = \frac{\mu_0 I_A I_B L}{2\pi d}$$
Step 3: Substitute the values and calculate.
$$F = \frac{4\pi \times 10^{-7} \times 8 \times 5 \times 0.1}{2\pi \times 0.04}$$
$$F = \frac{2 \times 10^{-7} \times 40 \times 0.1}{0.04}$$
$$F = 2 \times 10^{-5} \text{ N}$$
Step 4: Determine direction.
Since currents are in the same direction, the force is attractive.
Final Answer: $2 \times 10^{-5} \text{ N}$ towards wire B.
Step 1: Identify given values.
$I_A = 8.0 \text{ A}$, $I_B = 5.0 \text{ A}$
Distance, $d = 4.0 \text{ cm} = 0.04 \text{ m}$
Length of section, $L = 10 \text{ cm} = 0.1 \text{ m}$
Step 2: Apply the formula for force between parallel wires.
$$F = \frac{\mu_0 I_A I_B L}{2\pi d}$$
Step 3: Substitute the values and calculate.
$$F = \frac{4\pi \times 10^{-7} \times 8 \times 5 \times 0.1}{2\pi \times 0.04}$$
$$F = \frac{2 \times 10^{-7} \times 40 \times 0.1}{0.04}$$
$$F = 2 \times 10^{-5} \text{ N}$$
Step 4: Determine direction.
Since currents are in the same direction, the force is attractive.
Final Answer: $2 \times 10^{-5} \text{ N}$ towards wire B.
NCERT Q8: 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 $\mathbf{B}$ inside the solenoid near its centre.
Answer:
Step 1: Identify given values.
Length, $L = 80 \text{ cm} = 0.8 \text{ m}$
Total turns, $N = 5 \times 400 = 2000 \text{ turns}$
Current, $I = 8.0 \text{ A}$
Step 2: Calculate turns per unit length.
Turns per unit length, $n = \frac{N}{L} = \frac{2000}{0.8} = 2500 \text{ turns/m}$.
Step 3: Apply the formula for magnetic field inside a solenoid.
$$B = \mu_0 n I$$
Step 4: Substitute and calculate.
$$B = (4\pi \times 10^{-7}) \times 2500 \times 8.0$$
$$B = 4 \times 3.14 \times 10^{-7} \times 20000$$
$$B \approx 2.51 \times 10^{-2} \text{ T}$$
Final Answer: The magnetic field is $2.51 \times 10^{-2} \text{ T}$.
Step 1: Identify given values.
Length, $L = 80 \text{ cm} = 0.8 \text{ m}$
Total turns, $N = 5 \times 400 = 2000 \text{ turns}$
Current, $I = 8.0 \text{ A}$
Step 2: Calculate turns per unit length.
Turns per unit length, $n = \frac{N}{L} = \frac{2000}{0.8} = 2500 \text{ turns/m}$.
Step 3: Apply the formula for magnetic field inside a solenoid.
$$B = \mu_0 n I$$
Step 4: Substitute and calculate.
$$B = (4\pi \times 10^{-7}) \times 2500 \times 8.0$$
$$B = 4 \times 3.14 \times 10^{-7} \times 20000$$
$$B \approx 2.51 \times 10^{-2} \text{ T}$$
Final Answer: The magnetic field is $2.51 \times 10^{-2} \text{ T}$.
NCERT Q9: 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 $30^\circ$ with the direction of a uniform horizontal magnetic field of 0.80 T. What is the magnitude of torque experienced by the coil?
Answer:
Step 1: Identify given values.
Side of square, $a = 10 \text{ cm} = 0.1 \text{ m}$
Area, $A = a^2 = 0.01 \text{ m}^2$
Turns, $N = 20$
Current, $I = 12 \text{ A}$
Angle, $\theta = 30^\circ$
Magnetic field, $B = 0.80 \text{ T}$
Step 2: Apply torque formula.
$$\tau = N I A B \sin\theta$$
Step 3: Substitute and calculate.
$$\tau = 20 \times 12 \times 0.01 \times 0.80 \times \sin(30^\circ)$$
$$\tau = 2.4 \times 0.80 \times 0.5$$
$$\tau = 0.96 \text{ N m}$$
Final Answer: The torque is $0.96 \text{ N m}$.
Step 1: Identify given values.
Side of square, $a = 10 \text{ cm} = 0.1 \text{ m}$
Area, $A = a^2 = 0.01 \text{ m}^2$
Turns, $N = 20$
Current, $I = 12 \text{ A}$
Angle, $\theta = 30^\circ$
Magnetic field, $B = 0.80 \text{ T}$
Step 2: Apply torque formula.
$$\tau = N I A B \sin\theta$$
Step 3: Substitute and calculate.
$$\tau = 20 \times 12 \times 0.01 \times 0.80 \times \sin(30^\circ)$$
$$\tau = 2.4 \times 0.80 \times 0.5$$
$$\tau = 0.96 \text{ N m}$$
Final Answer: The torque is $0.96 \text{ N m}$.
NCERT Q10: Two moving coil meters, $M_1$ and $M_2$ have the following particulars:
$R_1 = 10\Omega, N_1=30, A_1=3.6 \times 10^{-3} \text{ m}^2, B_1=0.25 \text{ T}$
$R_2 = 14\Omega, N_2=42, A_2=1.8 \times 10^{-3} \text{ m}^2, B_2=0.50 \text{ T}$
(The spring constants are identical for the two meters). Determine the ratio of (a) current sensitivity and (b) voltage sensitivity of $M_2$ and $M_1$.
Answer:
Step 1: State the relevant formulas.
Current Sensitivity $I_s = \frac{N A B}{k}$
Voltage Sensitivity $V_s = \frac{N A B}{k R} = \frac{I_s}{R}$
Step 2: Calculate (a) Ratio of Current Sensitivity ($I_{s2}/I_{s1}$).
$$\frac{I_{s2}}{I_{s1}} = \frac{N_2 A_2 B_2}{N_1 A_1 B_1}$$
$$\frac{I_{s2}}{I_{s1}} = \frac{42 \times 1.8 \times 10^{-3} \times 0.50}{30 \times 3.6 \times 10^{-3} \times 0.25}$$
$$\frac{I_{s2}}{I_{s1}} = \frac{42}{30} \times \frac{1.8}{3.6} \times \frac{0.50}{0.25} = 1.4 \times 0.5 \times 2 = 1.4$$
Step 3: Calculate (b) Ratio of Voltage Sensitivity ($V_{s2}/V_{s1}$).
$$\frac{V_{s2}}{V_{s1}} = \frac{I_{s2}/R_2}{I_{s1}/R_1} = \frac{I_{s2}}{I_{s1}} \times \frac{R_1}{R_2}$$
$$\frac{V_{s2}}{V_{s1}} = 1.4 \times \frac{10}{14} = 1$$
Final Answer: Ratio of current sensitivity is $1.4$, and voltage sensitivity is $1$.
Step 1: State the relevant formulas.
Current Sensitivity $I_s = \frac{N A B}{k}$
Voltage Sensitivity $V_s = \frac{N A B}{k R} = \frac{I_s}{R}$
Step 2: Calculate (a) Ratio of Current Sensitivity ($I_{s2}/I_{s1}$).
$$\frac{I_{s2}}{I_{s1}} = \frac{N_2 A_2 B_2}{N_1 A_1 B_1}$$
$$\frac{I_{s2}}{I_{s1}} = \frac{42 \times 1.8 \times 10^{-3} \times 0.50}{30 \times 3.6 \times 10^{-3} \times 0.25}$$
$$\frac{I_{s2}}{I_{s1}} = \frac{42}{30} \times \frac{1.8}{3.6} \times \frac{0.50}{0.25} = 1.4 \times 0.5 \times 2 = 1.4$$
Step 3: Calculate (b) Ratio of Voltage Sensitivity ($V_{s2}/V_{s1}$).
$$\frac{V_{s2}}{V_{s1}} = \frac{I_{s2}/R_2}{I_{s1}/R_1} = \frac{I_{s2}}{I_{s1}} \times \frac{R_1}{R_2}$$
$$\frac{V_{s2}}{V_{s1}} = 1.4 \times \frac{10}{14} = 1$$
Final Answer: Ratio of current sensitivity is $1.4$, and voltage sensitivity is $1$.
NCERT Q11: In a chamber, a uniform magnetic field of $6.5 \text{ G} (1 \text{ G} = 10^{-4} \text{ T})$ is maintained. An electron is shot into the field with a speed of $4.8 \times 10^6 \text{ m/s}$ normal to the field. Explain why the path of the electron is a circle. Determine the radius of the circular orbit.
$(e=1.6 \times 10^{-19} \text{ C}, m_e=9.1 \times 10^{-31} \text{ kg})$
Answer:
Step 1: Explain why the path is circular.
Why a circular path? The magnetic force acts perpendicular to both the velocity of the electron and the magnetic field. This force provides the necessary centripetal force for circular motion without changing the speed.
Step 2: Radius Calculation.
Centripetal force = Magnetic Lorentz force
$$\frac{m v^2}{r} = e v B \implies r = \frac{m v}{e B}$$
Step 3: Substitute and calculate.
$$r = \frac{9.1 \times 10^{-31} \times 4.8 \times 10^6}{1.6 \times 10^{-19} \times 6.5 \times 10^{-4}}$$
$$r = \frac{43.68 \times 10^{-25}}{10.4 \times 10^{-23}}$$
$$r = 4.2 \times 10^{-2} \text{ m} = 4.2 \text{ cm}$$
Final Answer: The radius is $4.2 \text{ cm}$.
Step 1: Explain why the path is circular.
Why a circular path? The magnetic force acts perpendicular to both the velocity of the electron and the magnetic field. This force provides the necessary centripetal force for circular motion without changing the speed.
Step 2: Radius Calculation.
Centripetal force = Magnetic Lorentz force
$$\frac{m v^2}{r} = e v B \implies r = \frac{m v}{e B}$$
Step 3: Substitute and calculate.
$$r = \frac{9.1 \times 10^{-31} \times 4.8 \times 10^6}{1.6 \times 10^{-19} \times 6.5 \times 10^{-4}}$$
$$r = \frac{43.68 \times 10^{-25}}{10.4 \times 10^{-23}}$$
$$r = 4.2 \times 10^{-2} \text{ m} = 4.2 \text{ cm}$$
Final Answer: The radius is $4.2 \text{ cm}$.
NCERT Q12: In Exercise 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:
Step 1: State the formula for frequency of revolution.
$$\nu = \frac{v}{2\pi r} = \frac{v}{2\pi (mv/eB)} = \frac{eB}{2\pi m}$$
Step 2: Substitute values and calculate.
$$\nu = \frac{1.6 \times 10^{-19} \times 6.5 \times 10^{-4}}{2 \times 3.14 \times 9.1 \times 10^{-31}}$$
$$\nu \approx 18.18 \times 10^6 \text{ Hz} = 18.18 \text{ MHz}$$
Step 3: Explain the dependence on speed.
Explanation: No, the frequency is independent of the speed of the electron. As speed increases, the radius of the circular path increases proportionally, keeping the time period and frequency constant (this is the principle of a cyclotron).
Final Answer: Frequency is $18.18 \text{ MHz}$, independent of speed.
Step 1: State the formula for frequency of revolution.
$$\nu = \frac{v}{2\pi r} = \frac{v}{2\pi (mv/eB)} = \frac{eB}{2\pi m}$$
Step 2: Substitute values and calculate.
$$\nu = \frac{1.6 \times 10^{-19} \times 6.5 \times 10^{-4}}{2 \times 3.14 \times 9.1 \times 10^{-31}}$$
$$\nu \approx 18.18 \times 10^6 \text{ Hz} = 18.18 \text{ MHz}$$
Step 3: Explain the dependence on speed.
Explanation: No, the frequency is independent of the speed of the electron. As speed increases, the radius of the circular path increases proportionally, keeping the time period and frequency constant (this is the principle of a cyclotron).
Final Answer: Frequency is $18.18 \text{ MHz}$, independent of speed.
NCERT Q13: (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 $60^\circ$ 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?
Answer:
(a) Counter Torque calculation:
Step 1: Identify given values.
$N = 30$
$r = 0.08 \text{ m}$
$I = 6.0 \text{ A}$
$B = 1.0 \text{ T}$
$\theta = 60^\circ$
Area $A = \pi r^2 = 3.14 \times (0.08)^2 = 0.0201 \text{ m}^2$
Step 2: Calculate torque applied by magnetic field.
$$\tau = N I A B \sin\theta$$
$$\tau = 30 \times 6.0 \times 0.0201 \times 1.0 \times \sin(60^\circ)$$
$$\tau = 180 \times 0.0201 \times 0.866 \approx 3.133 \text{ N m}$$
To prevent the coil from turning, an equal and opposite counter torque of $3.13 \text{ N m}$ must be applied.
(b) Explanation for shape change:
No, the answer will not change. The torque depends on the total area enclosed by the coil, not on its shape, as long as the area $A$ remains the same.
(a) Counter Torque calculation:
Step 1: Identify given values.
$N = 30$
$r = 0.08 \text{ m}$
$I = 6.0 \text{ A}$
$B = 1.0 \text{ T}$
$\theta = 60^\circ$
Area $A = \pi r^2 = 3.14 \times (0.08)^2 = 0.0201 \text{ m}^2$
Step 2: Calculate torque applied by magnetic field.
$$\tau = N I A B \sin\theta$$
$$\tau = 30 \times 6.0 \times 0.0201 \times 1.0 \times \sin(60^\circ)$$
$$\tau = 180 \times 0.0201 \times 0.866 \approx 3.133 \text{ N m}$$
To prevent the coil from turning, an equal and opposite counter torque of $3.13 \text{ N m}$ must be applied.
(b) Explanation for shape change:
No, the answer will not change. The torque depends on the total area enclosed by the coil, not on its shape, as long as the area $A$ remains the same.
Extra Important Questions for 2026 Exams
These questions are specifically curated keeping the CBSE Board Exam Questions 2026 patterns in mind.
Important MCQ 1: A charged particle moves through a magnetic field perpendicular to its direction. What happens to its kinetic energy?
a) Increases
b) Decreases
c) Remains constant
d) Becomes zero
Answer: (c) Remains constant.
Explanation: The magnetic force is always perpendicular to velocity; hence no work is done.
Explanation: The magnetic force is always perpendicular to velocity; hence no work is done.
Important MCQ 2: Which instrument is used to measure the magnetic field?
a) Voltmeter
b) Ammeter
c) Galvanometer
d) Gaussmeter / Magnetometer
Answer: (d) Gaussmeter.
Important MCQ 3: If the number of turns in a moving coil galvanometer is doubled, its voltage sensitivity:
a) Doubles
b) Halves
c) Remains the same
d) Quadruples
Answer: (c) Remains the same.
Explanation: Resistance also doubles when turns are doubled.
Explanation: Resistance also doubles when turns are doubled.
Important MCQ 4: The SI unit of magnetic dipole moment is:
a) $\text{Am}^{2}$
b) $\text{Am}$
c) $\text{Tm/A}$
d) $\text{T/A}$
Answer: (a) $\text{Am}^{2}$
Important MCQ 5: Two parallel wires carrying current in opposite directions will:
a) Attract each other
b) Repel each other
c) Cancel each other's fields perfectly
d) No effect
Answer: (b) Repel each other.
Short Answer 6: Define Ampere based on the force between two parallel current-carrying conductors.
Answer: One Ampere is the value of steady current which, when maintained in each of the two very long, straight, parallel conductors of negligible cross-section placed one meter apart in a vacuum, would produce a force equal to $2 \times 10^{-7}$ Newtons per meter of length.
Difficulty: Moderate
Difficulty: Moderate
Short Answer 7: Why is a radial magnetic field required in a moving coil galvanometer?
Answer: A radial magnetic field ensures that the plane of the coil is always parallel to the magnetic field, regardless of its orientation. This makes $\sin\theta = 1$, keeping the torque ($\tau = N I A B$) maximum and producing a linear scale for the galvanometer.
Difficulty: Moderate
Difficulty: Moderate
Short Answer 8: State Biot-Savart Law in vector form.
Answer: $d\mathbf{B} = \frac{\mu_0}{4\pi} \frac{I (d\mathbf{l} \times \mathbf{r})}{r^3}$. It states that the magnetic field due to a current element is directly proportional to current, element length, and sine of the angle between them, and inversely proportional to the square of the distance.
Difficulty: Easy
Difficulty: Easy
Long Answer 9: Derive an expression for the magnetic field at a point on the axis of a circular current-carrying loop. Plot a graph showing the variation of the magnetic field along the axis.
Answer Hint: Use Biot-Savart Law. Resolve $d\mathbf{B}$ into perpendicular ($dB\cos\theta$) and parallel ($dB\sin\theta$) components. Perpendicular components cancel out. Integrate $dB\sin\theta$ over the loop.
Result: $$B = \frac{\mu_0 I R^2}{2(R^2 + x^2)^{3/2}}$$
Graph is a bell-shaped curve peaking at the center.
Result: $$B = \frac{\mu_0 I R^2}{2(R^2 + x^2)^{3/2}}$$
Graph is a bell-shaped curve peaking at the center.
Long Answer 10: How do you convert a Galvanometer into (a) an Ammeter and (b) a Voltmeter? Derive the necessary formulas.
Answer Hint:
(a) Connect a low resistance (shunt, $S$) in parallel. Formula: $$S = \frac{I_g G}{I - I_g}$$
(b) Connect a high resistance ($R$) in series. Formula: $$R = \frac{V}{I_g} - G$$
(a) Connect a low resistance (shunt, $S$) in parallel. Formula: $$S = \frac{I_g G}{I - I_g}$$
(b) Connect a high resistance ($R$) in series. Formula: $$R = \frac{V}{I_g} - G$$
Long Answer 11: Derive the expression for the force per unit length between two infinitely long straight parallel wires carrying currents in the same direction. Hence define one ampere.
Answer Hint: Use $B_1 = \frac{\mu_0 I_1}{2\pi d}$ from Ampere's Law. Apply $F = I_2 L B_1$.
Result: $$\frac{F}{L} = \frac{\mu_0 I_1 I_2}{2\pi d}$$
Result: $$\frac{F}{L} = \frac{\mu_0 I_1 I_2}{2\pi d}$$
Assertion-Reason 12:
Assertion: A proton and an alpha particle enter a uniform magnetic field with the same velocity. The radius of the alpha particle's path is twice that of the proton.
Reason: Radius $r = \frac{mv}{qB}$. Alpha particle has 4 times the mass and 2 times the charge of a proton.
(A = Both true, R is correct explanation. B = Both true, R is not correct explanation. C = A true, R false. D = A false, R true)
Answer: (A) Both are true, and R is the correct explanation.
Assertion-Reason 13:
Assertion: Cyclotron cannot accelerate electrons.
Reason: Electrons have very small mass, so they reach relativistic speeds quickly, throwing them out of phase with the oscillating electric field.
(A = Both true, R is correct explanation. B = Both true, R is not correct explanation. C = A true, R false. D = A false, R true)
Answer: (A) Both are true, and R is the correct explanation.
Case-Based 14: Moving Coil Galvanometer
A moving coil galvanometer operates on the principle that a current-carrying coil placed in a magnetic field experiences a torque. The restoring torque is provided by a phosphor-bronze hairspring.
Q1. What is the material of the core used inside the coil and why?
(Answer: Soft iron core, to increase the strength of the magnetic field and make it radial).
Q2. What is figure of merit?
(Answer: The current required to produce unit deflection).
(Answer: Soft iron core, to increase the strength of the magnetic field and make it radial).
Q2. What is figure of merit?
(Answer: The current required to produce unit deflection).
Case-Based 15: Lorentz Force
A charge $q$ moving with velocity $\mathbf{v}$ in a region where both electric field $\mathbf{E}$ and magnetic field $\mathbf{B}$ exist experiences a Lorentz force.
Q1. Under what condition will the particle pass straight through the fields undeflected?
(Answer: When electric force and magnetic force are equal and opposite, $v = E/B$. This is a velocity selector).
(Answer: When electric force and magnetic force are equal and opposite, $v = E/B$. This is a velocity selector).
Common Mistakes Students Make
- Forgetting the Vector Nature: Students often forget that magnetic field and force are vectors. Always specify the direction using the Right-Hand Thumb Rule or Fleming's Left-Hand Rule.
- Confusing Formulas: Mixing up the magnetic field at the center of a circular coil ($B = \frac{\mu_0 I}{2R}$) with the field of a straight wire ($B = \frac{\mu_0 I}{2\pi R}$). Note the $\pi$!
- Ammeter vs. Voltmeter Conversions: Remember: Ammeter conversion requires a parallel shunt (low resistance). Voltmeter requires a series resistance (high resistance).
- Angle in Torque Formula: In $\tau = N I A B \sin\theta$, $\theta$ is the angle between the normal to the area vector and the magnetic field, NOT the plane of the coil.
- Units Errors: Forgetting to convert centimeters or millimeters to meters before plugging values into formulas.
Exam Preparation Tips
- Master the Right-Hand Rules: Practice determining directions. 80% of mistakes in this chapter are direction-related.
- Formula Sheet: Create a single page with all formulas from this chapter. Write them down twice a week.
- Derivations are Key: Board exams frequently ask for the derivation of Ampere's Law applications, Biot-Savart axis application, and Galvanometer conversion. Practice writing them step-by-step.
- Time Management: In the board exam, do not spend more than 5 minutes on a 3-mark numerical. If stuck, write the formula, plug in given values, and move on.
FAQ Section
Is Moving Charges and Magnetism important for CBSE Class 12 boards?
Yes, Chapter 4 is extremely important. It carries a heavy weightage along with Chapter 5 (Magnetism and Matter) and forms the foundation for understanding electromagnetic induction.
Where can I download NCERT Class 12 Physics Solutions PDF?
You can download the chapter-wise PDF notes and NCERT solutions directly from educational portals like examspark.in to study offline without any distractions.
Which questions are most important from Chapter 4 Moving Charges and Magnetism?
Numerical problems based on Biot-Savart law applications, cyclotron frequency, parallel wire force derivation, and galvanometer conversion questions are the most frequently asked in board exams.
What is Lorentz Force?
Lorentz force is the combined force experienced by a charged particle moving through a region containing both electric and magnetic fields.
How to calculate the direction of the magnetic field?
You can use the Right-Hand Thumb Rule. If you point your right thumb in the direction of the electric current, your curled fingers will indicate the direction of the magnetic field lines.
Chapter 4, Moving Charges and Magnetism, can seem challenging at first because of the 3D visualizations and right-hand rules, but once you grasp the core concepts, it's highly scoring! Ratta mat maro (don't just memorize)—understand the logic behind Biot-Savart and Ampere's law. Bina tension ke exam phodna hai! Revise these updated NCERT solutions regularly, practice the board-style questions provided above, and download your quick revision notes from examspark.in. Keep practicing past year questions (PYQs) and step into the examination hall with 100% confidence. All the best for your 2026 board exams!
More Class 12 Physics Chapters
Ch 1: Electric Charges and Fields
Ch 2: Electrostatic Potential and Capacitance
Ch 3: Current Electricity
Ch 4: Moving Charges and Magnetism
Ch 5: Magnetism and Matter
Ch 6: Electromagnetic Induction
Ch 7: Alternating Current
Ch 8: Electromagnetic Waves
Ch 9: Ray Optics
Ch 10: Wave Optics
Ch 11: Dual Nature of Radiation and Matter
Ch 12: Atoms
Ch 13: Nuclei
Class 12 Mock QUIZs
Ch 14: Semiconductor Electronics