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HC Verma Solutions Class 12 Chapter 35 - Magnetic Field due to a Current

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Summary of HC Verma Solutions Part 2 Chapter 35: Magnetic Field Due To Current

This chapter talks about the bio-savant law and its application in Physics. This chapter further explains the right-hand thumb rule for shaping the current direction and calculation of force between parallels. Lastly, this chapter includes magnetic fields at points.


You can access the HC Verma Solutions for Chapter 35 - Magnetic Field Due To Current in PDF format for free. This means that you have the flexibility to study anytime and from anywhere, ensuring that you have the necessary materials at your fingertips.


The Class 12 HC Verma Solutions Magnetic Field Due To Current PDF provided by Vedantu is designed to support your learning journey. It is created by expert Physics teachers who possess a deep understanding of the concepts covered in the chapter. By offering solutions for all the exercises in the chapter, you have the opportunity to practice problem-solving in various contexts.


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Benefits of using Vedantu’s Class 12 HC Verma Solutions for Chapter 35 - Magnetic Field Due To Current

  • Solutions provided by expert Physics teachers with a deep understanding of the concepts.

  • Covers all exercises in the chapter for comprehensive practice.

  • Available in a free PDF format for easy accessibility anytime, anywhere.

  • Clear and concise explanations of the solutions to the exercises.


HC Verma Volume 2 Solutions Other Chapters:


To make the most of the HC Verma Chapter 35 - Magnetic Field Due To Current Solutions, Vedantu suggests the following tips:

Begin by thoroughly reading the chapter: It is important to have a solid understanding of the fundamental concepts and terminology before delving into the solutions.

Approach the examples step-by-step: Instead of simply memorizing the solutions, focus on understanding the logic and reasoning behind each step. This will enhance your comprehension and enable you to tackle similar problems.

Attempt the illustrative exercises independently: Challenge yourself to solve the exercises on your own before referring to the solutions. In case you encounter difficulties, the solutions can provide guidance, but attempting the problems independently first will improve your problem-solving skills.

Practice diligently: Engage in regular and consistent practice to enhance your proficiency in solving physics problems. The more you practice, the more proficient you will become.

In summary, Vedantu's Class 12 HC Verma Solutions for Chapter 35 - Magnetic Field Due To Current offer expertly crafted solutions provided in a convenient and accessible PDF format. Utilizing these solutions, along with the suggested study tips, will empower you to excel in your physics studies.


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FAQs on HC Verma Solutions Class 12 Chapter 35 - Magnetic Field due to a Current

1. What are the fundamental laws and formulas from Chapter 35, 'Magnetic Field due to a Current', that are essential for solving HC Verma problems?

To solve the problems in HC Verma's Chapter 35, you must have a strong grasp of two key principles. The solutions primarily rely on:

  • Biot-Savart Law: This law is used to calculate the magnetic field produced by a small current element. The formula is dB = (μ₀/4π) * (I dl × ) / r². It is a fundamental law used for current distributions of any shape.

  • Ampere's Circuital Law: This law relates the line integral of the magnetic field around a closed loop to the net current passing through that loop. The formula is ∮B⋅dl = μ₀I_enclosed. It is most effective for situations with high symmetry, such as long wires, solenoids, and toroids.

Mastering the application of these two laws is crucial for successfully navigating the exercises in this chapter.

2. How should the Right-Hand Thumb Rule be applied correctly to determine the magnetic field direction for a circular loop as per HC Verma exercises?

For a current-carrying circular loop, the Right-Hand Thumb Rule is applied slightly differently than for a straight wire. To find the direction of the magnetic field at the centre of the loop, you should curl the fingers of your right hand in the direction of the current flow in the loop. Your outstretched thumb will then point in the direction of the magnetic field. For example, if the current is flowing in an anti-clockwise direction, the magnetic field at the centre will point outwards, perpendicular to the plane of the loop.

3. What is the step-by-step method to calculate the magnetic field on the axis of a current-carrying circular coil using Biot-Savart Law?

The solutions in HC Verma for this problem follow a systematic approach:

  1. First, consider a small current element Idl on the circumference of the coil.

  2. Determine the position vector r from this element to the point P on the axis where the field is to be calculated.

  3. Apply the Biot-Savart Law to find the magnetic field dB due to this element. The direction of dB will be perpendicular to the plane containing Idl and r.

  4. Resolve the vector dB into two components: one along the axis of the coil and another perpendicular to it.

  5. Due to symmetry, the perpendicular components from all elements around the loop will cancel each other out.

  6. Finally, integrate the axial components over the entire length of the circular coil to get the net magnetic field. This results in the formula B = (μ₀NIR²) / 2(x² + R²)^(3/2).

4. When solving problems in Chapter 35, how does one decide whether to use Biot-Savart Law or Ampere's Circuital Law?

The choice between these two laws depends entirely on the symmetry of the current distribution. You should use:

  • Biot-Savart Law for current configurations that lack symmetry, such as finding the field from a finite wire at an arbitrary point, a segment of a circular arc, or any irregular shape. It is a more universal tool but often involves complex integration.

  • Ampere's Circuital Law for highly symmetrical current configurations where you can easily define an Amperian loop over which the magnetic field is constant in magnitude. This is ideal for an infinitely long straight wire, a long solenoid, and a toroid. It simplifies the calculation significantly in these specific cases.

5. What is a common misconception about the magnetic field inside an ideal solenoid, and how do HC Verma solutions address this?

A common misconception is that the magnetic field strength varies across the cross-section inside a solenoid. HC Verma solutions are based on the ideal model, which clarifies that for a long solenoid, the magnetic field inside is strong, uniform, and parallel to its axis, especially away from the ends. The field lines are straight and evenly spaced, indicating uniformity. The magnetic field outside the solenoid is considered negligible or zero in most problems. Understanding this uniformity is key to applying Ampere's law correctly to find B = μ₀nI.

6. Why is the magnetic field at the end of a very long solenoid exactly half of the field at its centre, a concept crucial for some HC Verma problems?

This is a direct consequence of the geometry of the setup when using the Biot-Savart law for integration. At the centre of an infinitely long solenoid, the field is integrated over an angular range from -90° to +90°. However, when calculating the field at one of its ends (treating it as a semi-infinite solenoid), the integration is performed over an angular range from 0° to +90°. This change in integration limits results in a value that is precisely one-half of the field at the centre. Therefore, if the field at the centre is B = μ₀nI, the field at the end is B = (1/2)μ₀nI.

7. How is the solution for the magnetic field inside a toroid different from that of a straight solenoid?

While both a solenoid and a toroid confine the magnetic field within their coils, the nature of the field differs slightly. In an ideal straight solenoid, the magnetic field is perfectly uniform inside. However, in a toroid (a solenoid bent into a circle), the magnetic field is not perfectly uniform across its cross-section. The field lines are concentric circles, and the strength of the magnetic field varies with the radial distance from the centre of the toroid, given by the formula B = (μ₀NI) / (2πr). This means the field is stronger on the inner side of the toroidal coil and weaker on the outer side, a detail important for precise calculations in HC Verma exercises.