RBSE Class 12 Physics Chapter 4 Solutions: Moving Charges and Magnetism – Master the Concepts!

🚀 Conquer Your Board Exams with Complete Solutions & Key Concepts

Are you an RBSE Class 12 student looking to ace your Physics exams? Chapter 4, Moving Charges and Magnetism, is one of the most critical and scoring chapters. This comprehensive guide provides you with detailed, step-by-step solutions to all NCERT textbook exercises (which form the basis for RBSE exams) and a quick recap of the most important formulas and concepts.

Stop struggling with complex derivations and tricky numericals! Our SEO-friendly solutions are designed to help you understand the why behind the what, ensuring you are prepared for both theoretical and numerical questions in the Rajasthan Board examination.

📖 Chapter 4: Moving Charges and Magnetism – Key Topics

Before diving into the solutions, here are the essential concepts you must master:

Topic	Key Concepts to Remember	Core Formula(s)
Magnetic Force	Magnetic force on a moving charge (Lorentz Force). Motion of a charged particle in a uniform magnetic field (circular path, helical path). Velocity selector.	F=q(v×B) F=qE+q(v×B) (Lorentz Force) r=qBmv​ (Radius of circular path)
Force on a Conductor	Magnetic force on a current-carrying conductor in a uniform magnetic field. Fleming’s Left-Hand Rule.	F=I(l×B)
Biot-Savart Law	Principle to find the magnetic field due to a current element. Magnetic field due to a circular current loop (at the center and on the axis).	dB=4πμ0​I​r3dl×r​ Bcentre​=2Rμ0​I​
Ampere’s Circuital Law	Relation between line integral of magnetic field and current enclosed. Application for a long straight wire, solenoid, and toroid.	∮B⋅dl=μ0​Ienclosed​ Bsolenoid​=μ0​nI
Torque on a Loop	Torque experienced by a rectangular current loop in a uniform magnetic field. Magnetic dipole moment of a current loop.	τ=M×B τ=NIABsinθ M=NIA (Magnetic Moment)
Moving Coil Galvanometer	Principle, construction, and conversion into Ammeter and Voltmeter. Current and Voltage Sensitivity.	Is​=Iϕ​=kNAB​ (Current Sensitivity)

✅ RBSE Class 12 Physics Chapter 4 – Detailed Solutions

Here are the step-by-step solutions to the most important numerical problems from your textbook.

💡 Example 1: Magnetic Field at the Centre of a Circular Coil (NCERT Exercise Q. 4.1)

Question: 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:

1. Given Data:
   • Number of turns, N=100
   • Radius of the coil, R=8.0 cm=8.0×10−2 m
   • Current, I=0.40 A
   • Permeability of free space, μ0​=4π×10−7 T m A−1
2. Formula Used: The magnetic field at the centre of a circular coil with N turns is: B=2Rμ0​NI​
3. Calculation: B=2×(8.0×10−2 m)(4π×10−7 T m A−1)×100×0.40 A​B=16×10−24π×10−7×40​ TB=16×10−2160π×10−7​ T=10π×10−5 TB≈3.14×10−4 T
4. Answer: The magnitude of the magnetic field at the centre of the coil is 3.14×10−4 T.

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💡 Example 2: Force on a Current-Carrying Wire (NCERT Exercise Q. 4.5)

Question: What is the magnitude of the 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:

1. Given Data:
   • Current, I=8 A
   • Magnetic field, B=0.15 T
   • Angle between wire and field, θ=30∘
2. Formula Used: The magnetic force on a current-carrying conductor of length l is F=IlBsinθ. We need the force per unit length, lF​. lF​=IBsinθ
3. Calculation: lF​=(8 A)×(0.15 T)×sin(30∘)lF​=1.2×21​ N/mlF​=0.6 N/m
4. Answer: The magnetic force per unit length on the wire is 0.6 N/m.

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💡 Example 3: Radius of Path of a Charged Particle (NCERT Exercise Q. 4.11)

Question: An electron is moving in a uniform magnetic field B (perpendicular to v). Show that its path is circular and derive the expression for the radius of the circular path.

Derivation & Solution (Radius part):

1. When a charge q moves with velocity v perpendicular to a uniform magnetic field B, the magnetic force acting on it is F=qvBsin(90∘)=qvB.
2. The magnetic force F is always perpendicular to the velocity v and the magnetic field B.
3. Since the force is perpendicular to the velocity, it changes only the direction of motion, not the speed. This constant force directed towards a center provides the necessary centripetal force.
4. Equating the magnetic force to the centripetal force: Fmagnetic​=Fcentripetal​qvB=rmv2​
5. Solving for the radius r: r=qBmv​

Answer: The radius of the circular path is r=qBmv​.

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🎯 Important Questions for RBSE Board Exam (Short & Long Answer)

For maximum marks in the RBSE board, focus on these derivations and concepts:

1. Biot-Savart Law: State the law and use it to derive the magnetic field on the axis of a circular current loop.
2. Ampere’s Circuital Law: State the law and apply it to find the magnetic field inside a long solenoid.
3. Lorentz Force: Write the expression for the Lorentz Force and explain the motion of a charged particle when v is (a) perpendicular to B, and (b) makes an angle θ with B.
4. Moving Coil Galvanometer (MCG): Explain its principle and derive the expression for the torque (τ=NIABsinα). Explain how it is converted into an Ammeter and a Voltmeter.
5. Force between two parallel current-carrying conductors: Derive the formula for the force per unit length and use it to define the SI unit of current, Ampere.

Chapter 4

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🚀 Get Ahead with Previous Year Questions (PYQs)

Practicing Rajasthan Board Previous Year Questions (PYQs) is the most effective way to guarantee a top rank. These questions often repeat or follow the same pattern.