Summary of HC Verma Solutions Part 2 Chapter 35: Magnetic Field Due To Current
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 × r̂) / 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:
First, consider a small current element Idl on the circumference of the coil.
Determine the position vector r from this element to the point P on the axis where the field is to be calculated.
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.
Resolve the vector dB into two components: one along the axis of the coil and another perpendicular to it.
Due to symmetry, the perpendicular components from all elements around the loop will cancel each other out.
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.
