Skip to main content
Super TutorResources
Chapter 7 of 14
QuestionsNotesFlashcardsSummaryFormulasQuizConceptsSyllabusPlan
NCERT Solutions

Moving Charges and Magnetism

CBSE · Class 12 · Physics

NCERT Solutions for Moving Charges and Magnetism — CBSE Class 12 Physics.

118 questions70 flashcards5 concepts

Interactive on Super Tutor

Studying Moving Charges and Magnetism? Get the full interactive chapter.

Quizzes, flashcards, AI doubt-solver and a step-by-step study plan — built for ncert solutions and more.

Try Super Tutor — Free

1,000+ Class 12 students started this chapter today

Illustrates the magnetic field produced by a circular current loop at a point on its axis, showing how perpendicular components of `dB` cancel and axial components add up.
Super Tutor

Super Tutor has 16+ illustrations like this for Moving Charges and Magnetism alone — flashcards, concept maps, and step-by-step visuals.

See them all
14 Questions Solved · 1 Section

7 worked solutions below. Unlock all 14 free in Super Tutor

EXERCISES

4.1A 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?Show solution
For a circular coil, magnetic field at the centre is

B=μ0NI2RB=\dfrac{\mu_0 N I}{2R}

Given: N=100N=100, I=0.40 AI=0.40\,\text{A}, R=8.0 cm=0.08 mR=8.0\,\text{cm}=0.08\,\text{m}.

B=4π×10−7×100×0.402×0.08 B=\frac{4\pi\times10^{-7}\times100\times0.40}{2\times0.08}

=16π×10−60.16=100π×10−6≈3.14×10−4 T =\frac{16\pi\times10^{-6}}{0.16} =100\pi\times10^{-6} \approx 3.14\times10^{-4}\,\text{T}

So the field at the centre is 3.14×10−4 T3.14\times10^{-4}\,\text{T}.

Not sure why a step works? check your working in Super Tutor

4.2A 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?Show solution
For a long straight wire,

B=μ0I2πrB=\dfrac{\mu_0 I}{2\pi r}

Given: I=35 AI=35\,\text{A}, r=20 cm=0.20 mr=20\,\text{cm}=0.20\,\text{m}.

B=4π×10−7×352π×0.20 B=\frac{4\pi\times10^{-7}\times35}{2\pi\times0.20}

=140×10−70.40=3.5×10−5 T =\frac{140\times10^{-7}}{0.40}=3.5\times10^{-5}\,\text{T}

Not sure why a step works? check your working in Super Tutor

4.3A 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.Show solution
Use the field due to a long straight wire:

B=μ0I2πrB=\dfrac{\mu_0 I}{2\pi r}

Given: I=50 AI=50\,\text{A}, r=2.5 mr=2.5\,\text{m}.

B=4π×10−7×502π×2.5 B=\frac{4\pi\times10^{-7}\times50}{2\pi\times2.5}

=200×10−75=4×10−6 T =\frac{200\times10^{-7}}{5}=4\times10^{-6}\,\text{T}

Direction: current is from north to south. At a point east of the wire, by the right-hand rule, the magnetic field is vertically upward.

So the field is 4.0×10−6 T4.0\times10^{-6}\,\text{T} upward.

Not sure why a step works? check your working in Super Tutor

4.4A 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?Show solution
For a long straight wire,

B=μ0I2πrB=\dfrac{\mu_0 I}{2\pi r}

Given: I=90 AI=90\,\text{A}, r=1.5 mr=1.5\,\text{m}.

B=4π×10−7×902π×1.5 B=\frac{4\pi\times10^{-7}\times90}{2\pi\times1.5}

=360×10−73=1.2×10−5 T =\frac{360\times10^{-7}}{3}=1.2\times10^{-5}\,\text{T}

Direction: current is east to west, and the point is below the wire. By the right-hand rule, the magnetic field is vertically downward.

Not sure why a step works? check your working in Super Tutor

4.5What is the magnitude of 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?Show solution
Magnetic force per unit length on a wire is

f=IBsin⁡θf=IB\sin\theta

Given: I=8 AI=8\,\text{A}, B=0.15 TB=0.15\,\text{T}, θ=30∘\theta=30^\circ.

f=8×0.15×sin⁡30∘ f=8\times0.15\times\sin30^\circ

=8×0.15×12=0.60 N m−1 =8\times0.15\times\frac12=0.60\,\text{N m}^{-1}

Not sure why a step works? check your working in Super Tutor

4.6A 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?Show solution
Magnetic force on a current-carrying wire is

F=IlBsin⁡θF=IlB\sin\theta

Here the wire is perpendicular to the solenoid axis, so it is perpendicular to BB and sin⁡θ=1\sin\theta=1.

Given: I=10 AI=10\,\text{A}, l=3.0 cm=0.03 ml=3.0\,\text{cm}=0.03\,\text{m}, B=0.27 TB=0.27\,\text{T}.

F=10×0.03×0.27=0.081 N F=10\times0.03\times0.27=0.081\,\text{N}

So the force is 8.1×10−2 N8.1\times10^{-2}\,\text{N}.

Not sure why a step works? check your working in Super Tutor

4.7Two 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.Show solution
Force between two long parallel currents is

F=μ0IaIb2πdL F=\frac{\mu_0 I_a I_b}{2\pi d}L

Given: Ia=8.0 AI_a=8.0\,\text{A}, Ib=5.0 AI_b=5.0\,\text{A}, d=4.0 cm=0.04 md=4.0\,\text{cm}=0.04\,\text{m}, L=10 cm=0.10 mL=10\,\text{cm}=0.10\,\text{m}.

F=4π×10−7×8×52π×0.04×0.10 F=\frac{4\pi\times10^{-7}\times8\times5}{2\pi\times0.04}\times0.10

=160×10−70.08×0.10=2.0×10−5 N =\frac{160\times10^{-7}}{0.08}\times0.10 =2.0\times10^{-5}\,\text{N}

Since the currents are in the same direction, the wires attract.

Not sure why a step works? check your working in Super Tutor

4.8A 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 B inside the solenoid near its centre.
4.9A 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° with the direction of a uniform horizontal magnetic field of magnitude 0.80 T. What is the magnitude of torque experienced by the coil?
4.10Determine the ratio of (a) current sensitivity and (b) voltage sensitivity of M₂ and M₁.
4.11In a chamber, a uniform magnetic field of 6.5 G (1 G = 10⁻⁴ T) is maintained. An electron is shot into the field with a speed of 4.8 × 10⁶ m s⁻¹ normal to the field. Explain why the path of the electron is a circle. Determine the radius of the circular orbit. (e = 1.5 × 10⁻¹⁹ C, mₑ = 9.1×10⁻³¹ kg)
4.12In Exercise 4.11 obtain the frequency of revolution of the electron in its circular orbit. Does the answer depend on the speed of the electron? Explain.
4.13(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° with the normal of the coil. Calculate the magnitude of the counter torque that must be applied to prevent the coil from turning.
4.13(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? (All other particulars are also unaltered.)

7 more solved questions in Moving Charges and Magnetism

Every remaining exercise is solved step by step in Super Tutor, plus practice quizzes and flashcards for this chapter. Free to start.

Stuck on a step?

Ask Super Tutor AI to explain any solution on this page in a simpler way — free, 24x7.

Ask a Doubt Free

Frequently Asked Questions

What are the important topics in Moving Charges and Magnetism for CBSE Class 12 Physics?
Moving Charges and Magnetism covers several key topics that are frequently asked in CBSE Class 12 board exams. Focus on the core concepts listed on this page and practise related questions to build confidence.
How to score full marks in Moving Charges and Magnetism — CBSE Class 12 Physics?
Understand the core concepts first, then work through the 118 practice questions available for this chapter. Revise formulas and definitions regularly, and use flashcards for quick recall before the exam.
Where can I get free NCERT Solutions for Moving Charges and Magnetism Class 12 Physics?
This page has free step-by-step NCERT Solutions for every exercise question in Moving Charges and Magnetism (CBSE Class 12 Physics) — written the way examiners award marks: given, formula, working, answer.

Sources & Official References

Content is aligned to the official syllabus. Refer to the board website for the latest curriculum.

More resources for Moving Charges and Magnetism

All 14 chapters — CBSE Class 12 Physics NCERT Solutions

For serious students

Get the full Moving Charges and Magnetism chapter — for free.

Quizzes, flashcards, AI doubt-solver and a step-by-step study plan for CBSE Class 12 Physics.

Try Super Tutor Free
Try the full chapter — Free