Class 10th Physics Chapter 15 Exercise Solved Numericals

Physics can seem daunting, but when approached with the right tools, it becomes an engaging exploration of natural phenomena.

Chapter 15 of Class 10 Physics focuses on Electromagnetism, a fascinating subject that delves into the interaction between electricity and magnetism.

In this article, we provide meticulously solved numerical problems from this chapter to help students master the concepts effectively. These solutions are derived from trusted academic resources, ensuring accuracy and reliability.

Topics Covered in These Notes

From pages 43 to 47 of the textbook, the following key topics are highlighted:

1. Magnetic effects of a steady current
   
2. Forces on current-carrying conductors in magnetic fields
   
3. Fleming’s Left-Hand Rule for determining force direction
   
4. Applications of electromagnetism, including motors and generators
   
5. Solenoids and their magnetic properties

These topics lay a strong foundation for understanding numerical problems in electromagnetism.

Numerical Problems

Numerical problems are a core component of physics, bridging theoretical concepts with practical applications. Below are examples of solved numerical problems covered in this section:

1. Calculation of Magnetic Field Strength
   Problem: A solenoid of length 50 cm has 500 turns and carries a current of 2 A. Calculate the magnetic field strength inside the solenoid.
   
   Solution:
   The magnetic field strength BBB is given by:B=μ0⋅n⋅IB = \mu_0 \cdot n \cdot IB=μ0​⋅n⋅IWhere:
   • μ0=4π×10−7 T\cdotpm/A\mu_0 = 4\pi \times 10^{-7} \, \text{T·m/A}μ0​=4π×10−7T\cdotpm/An=Number of turnsLength of solenoid=5000.5=1000 turns/mn = \frac{\text{Number of turns}}{\text{Length of solenoid}} = \frac{500}{0.5} = 1000 \, \text{turns/m}n=Length of solenoidNumber of turns​=0.5500​=1000turns/mI=2 AI = 2 \, \text{A}I=2A
   
   Substituting values:B=(4π×10−7)⋅1000⋅2B = (4\pi \times 10^{-7}) \cdot 1000 \cdot 2 B=(4π×10−7)⋅1000⋅2 B=2.51×10−3 TB = 2.51 \times 10^{-3} \, \text{T} B=2.51×10−3TAnswer: The magnetic field strength inside the solenoid is 2.51 mT2.51 \, \text{mT}2.51mT.
   
2. Force on a Conductor
   Problem: A straight conductor of length 2 m carrying a current of 5 A is placed in a uniform magnetic field of 0.1 T0.1 \, \text{T}0.1T. Find the force acting on the conductor if the magnetic field is perpendicular to the conductor.
   
   Solution:
   The force FFF is given by:F=I⋅L⋅BF = I \cdot L \cdot BF=I⋅L⋅BWhere:
   • I=5 AI = 5 \, \text{A}I=5AL=2 mL = 2 \, \text{m}L=2mB=0.1 TB = 0.1 \, \text{T}B=0.1T
   Substituting values:F=5⋅2⋅0.1=1 NF = 5 \cdot 2 \cdot 0.1 = 1 \, \text{N}F=5⋅2⋅0.1=1NAnswer: The force acting on the conductor is 1 N1 \, \text{N}1N.

Tool for Success in Exams

Understanding and practicing numerical problems are essential for excelling in physics exams. These problems enhance critical thinking and problem-solving skills, ensuring students are well-prepared.

The solved examples provided here are designed to build confidence and a deep understanding of electromagnetism principles.

Colored Notes

To make learning easier and visually engaging, the notes for Chapter 15 include color-coded diagrams and highlighted formulas. These features are invaluable for understanding complex concepts such as:

1. The direction of magnetic fields (illustrated using Fleming’s Left-Hand Rule diagrams)
   
2. The working of solenoids, motors, and generators, with detailed labeled schematics
   
3. Step-by-step derivations and calculations of key formulas, such as F=I⋅L⋅BF = I \cdot L \cdot BF=I⋅L⋅B and B=μ0⋅n⋅IB = \mu_0 \cdot n \cdot IB=μ0​⋅n⋅I

These visual aids not only enhance retention but also make revision faster and more effective.

Notes Are Free to Use

One of the most significant advantages of these notes is their availability at no cost. They are accessible to all students, ensuring equal opportunities for academic success.

By providing free access, these resources empower students to overcome barriers and focus solely on their learning journey.

Notes Are Mistake-Free

Accuracy is critical when it comes to academic resources, and these notes uphold the highest standards. Each solution has been carefully checked and cross-verified by subject matter experts.

This ensures that students can rely on these notes with confidence, avoiding the frustration of encountering errors during their study sessions.

Chapter 15 Class 10th Physics Notes

1. Class 10th Physics Chapter 15 Notes (MCQs, Short & Long Questions) 
2. Class 10th Physics Chapter 15 Solved Exercise (MCQs, Short & Long Questions)

Conclusion

Chapter 15 of Class 10 Physics offers a captivating exploration of electromagnetism, a topic that connects science with real-world applications.

The solved numerical problems and comprehensive notes presented here aim to simplify this subject, making it approachable and understandable for students.

By focusing on practical problem-solving and providing error-free, visually engaging content, these notes are a perfect tool for mastering the concepts of electromagnetism.

FAQs

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