NCERT Solutions for Class 12 Physics Chapter 4: Moving Charges and Magnetism – Free PDF Download

Complete, verified solutions to the NCERT Class 12 Physics Chapter 4 (Moving Charges and Magnetism) exercise questions. These Class 12 Physics Chapter 4 solutions are also useful as quick revision notes before exams.

NCERT Exercise Solutions

4.1 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?
Ans: B=μ₀NI/2R=(4π×10⁻⁷)(100)(0.40)/(2×0.08)=3.14×10⁻⁴ T.

4.2 A long straight wire carries a current of 35A. What is the magnitude of the field B at a point 20cm from the wire?
Ans: B=μ₀I/2πr=(4π×10⁻⁷)(35)/(2π×0.20)=3.5×10⁻⁵ T.

4.3 A long straight wire in the horizontal plane carries a current of 50A in north to south direction. Give the magnitude and direction of B at a point 2.5m east of the wire.
Ans: B=μ₀I/2πr=(4π×10⁻⁷)(50)/(2π×2.5)=4×10⁻⁲ T, directed vertically upward (by the right-hand rule: point the right thumb south along the current, and at a point to the east the curling fingers point upward).

4.4 A horizontal overhead power line carries a current of 90A in east to west direction. What is the magnitude and direction of B due to the current 1.5m below the line?
Ans: B=μ₀I/2πr=(4π×10⁻⁷)(90)/(2π×1.5)=1.2×10⁻⁵ T, directed horizontally toward the south (by the right-hand rule: point the right thumb west along the current, and at a point directly below, the curling fingers point toward the south).

4.5 A straight wire carrying a current of 12A is bent into a semicircular arc of radius 2.0cm. What is B at the centre?
Ans: For a semicircular arc: B=μ₀I/4R=(4π×10⁻⁷)(12)/(4×0.02)=1.9×10⁻⁴ T. (The two straight portions on either side, if along lines through the centre, contribute zero field.)

4.6 A 100-turn closely wound circular coil of radius 8.0cm carries a current of 0.40A. What is the magnitude of the magnetic moment associated with the coil?
Ans: m=NIA=100×0.40×π(0.08)²=100×0.40×0.0201=0.80 A·m².

4.7 A closely wound solenoid of 2000 turns and area of cross-section 1.6×10⁻⁴ m² carries a current of 4.0A. What is the magnitude of the magnetic moment?
Ans: m=NIA=2000×4.0×1.6×10⁻⁴=1.28 A·m².

4.8 A galvanometer coil has resistance 12Ω and the metre shows full-scale deflection for a current of 3mA. How will you convert it into a voltmeter of range 0 to 18V?
Ans: Need series resistance R such that V=Ig(G+R): 18=0.003(12+R) ⇒ 6000=12+R ⇒ R=5988Ω.

4.9 A galvanometer coil has resistance 15Ω and gives full-scale deflection for a current of 4mA. How will you convert it into an ammeter of range 0 to 6A?
Ans: Shunt S needed: IgG=(I−Ig)S ⇒ 0.004×15=(6−0.004)×S ⇒ S=0.06/5.996≈0.01Ω.

Frequently Asked Questions

What is the magnetic field formula at the centre of a circular coil?
B=μ₀NI/2R.

What is the field due to a semicircular arc at its centre?
B=μ₀I/4R, half of the full-circle formula.

Written by Satish

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