Important Derivations for Class 12 Physics Chapter 4 Moving Charges and Magnetism

Important Derivations for Class 12 Physics Chapter 4 Moving Charges and Magnetism

Here we are providing important derivations for Class 12 Physics Chapter 4 Moving Charges and Magnetism.

(1) Biot Savarts Law

With the help of experimental results, Biot and Savart arrived at a mathematical expression that gives
the magnetic field at some point in space in terms of the current that produces the field.

Experimentally, it was found that, magnetic field at point P varies as:

(1) $$\mathrm{dB} \propto \mathrm{id} \ell \sin \theta$$ (2) $$\mathrm{dB} \propto \frac{1}{\mathrm{r}^2}$$ $$ \begin{aligned} & \Rightarrow \mathrm{dB} \propto \frac{\mathrm{id} \ell \sin \theta}{\mathrm{r}^2} \\ & \mathrm{~dB}=\frac{\mu_0}{4 \pi} \frac{\mathrm{i} \ell \sin \theta}{\mathrm{r}^2} \end{aligned} $$ here $$ \mu_0=4 \pi \times 10^{-7} \frac{\mathrm{Tm}}{\mathrm{A}}$$ (permeability of free space)
Unit of B: T or Wb/m²
Scalar Form $$ \mathrm{dB}=\frac{\mu_0}{4 \pi} \frac{\mathrm{id} \ell \sin \theta}{\mathrm{r}^2} $$ Vector Form $$ \mathrm{d} \overrightarrow{\mathrm{B}}=\frac{\mu_0}{4 \pi} \frac{\mathrm{i}(\mathrm{d} \vec{\ell} \times \mathrm{r})}{\mathrm{r}^3} $$

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(2) The parallel coaxial circular coils of equal radius R and equal number of turns N carry equal currents I in the same direction and are separated by a distance 2 R. Find the magnitude and direction of the net magnetic field produced at the mid-point of the line joining their centres.

To find the magnitude and direction of the net magnetic field at the midpoint of the line joining the centers of two parallel coaxial circular coils, we can use the principle of superposition.

Let’s consider the two circular coils, Coil 1 and Coil 2, carrying equal currents \(I\) in the same direction. They have equal radii \(R\) and are separated by a distance of \(2R\). The midpoint of the line joining their centers is the point where we want to find the net magnetic field.

At the midpoint, the magnetic field produced by Coil 1 and Coil 2 will add together to give the net magnetic field.

The magnetic field (\(B_1\)) produced at the midpoint due to Coil 1 can be calculated using the formula for the magnetic field at the center of a circular coil:

\[ B_1 = \frac{{\mu_0 \cdot N \cdot I}}{{2R}} \] where: \( \mu_0 \) = Permeability of free space (\(4\pi \times 10^{-7} \, \text{T}\cdot\text{m/A}\)),
\( N \) = Number of turns in each coil (since they have equal turns, \(N\) is the same for both coils),
\( I \) = Current flowing through each coil,
\( R \) = Radius of each coil.

The magnetic field (\(B_2\)) produced at the midpoint due to Coil 2 is the same as \(B_1\) since both coils are identical and carry the same current.

Now, to find the net magnetic field (\(B_{\text{net}}\)) at the midpoint, we add the contributions from both coils: \[ B_{\text{net}} = B_1 + B_2 \] \[ B_{\text{net}} = \frac{{\mu_0 \cdot N \cdot I}}{{2R}} + \frac{{\mu_0 \cdot N \cdot I}}{{2R}} \] \[ B_{\text{net}} = \frac{{2 \mu_0 \cdot N \cdot I}}{{2R}} \] \[ B_{\text{net}} = \frac{{\mu_0 \cdot N \cdot I}}{{R}} \] So, the magnitude of the net magnetic field at the midpoint is \( \frac{{\mu_0 \cdot N \cdot I}}{{R}} \). The direction of the net magnetic field is along the axis of the coils, which is the line joining their centers. It points from Coil 1 to Coil 2.

Therefore, the magnitude of the net magnetic field is \( \frac{{\mu_0 \cdot N \cdot I}}{{R}} \) and its direction is along the axis of the coils from Coil 1 to Coil 2.

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Why students face difficulty in physics derivations?

There can be several reasons why students face difficulty in physics derivations, including:

1. Lack of foundation: A lack of understanding of the underlying concepts and principles can make it difficult to follow the logical steps in a derivation.
2. Mathematical background: Physics often involves complex mathematical calculations, and students who struggle with math may find it challenging to perform the necessary calculations.
3. Limited practice: Regular practice is key to developing the skills required for physics derivations, and students who have limited practice opportunities may struggle.
4. Poor problem-solving skills: Physics problems often require creative problem-solving skills, and students who struggle with this aspect of physics may find derivations particularly challenging.
5. Limited exposure to different problems: Physics derivations can vary widely in their level of complexity, and students who have limited exposure to different types of problems may struggle when faced with a new challenge.
6. Confusion with notation: Physics often uses a specialized notation, and students who are unfamiliar with this notation may struggle to follow the steps in a derivation.

Remembering physics derivations can be a challenge, but here are some tips that may help:

1. Practice, practice, practice: Regularly practicing physics derivations helps to build muscle memory and improve recall.
2. Visualize the process: Try to visualize the steps involved in a derivation, as well as the physical meanings behind each equation.
3. Make connections: Try to relate each step of the derivation to a concept or formula you already know, which can help to reinforce your understanding.
4. Write it down: Writing out a derivation helps to solidify your understanding and makes it easier to remember.
5. Understand the physical meaning: Try to understand the physical meaning behind each equation, which can help you to remember the derivation in a broader context.
6. Teach someone else: Teaching someone else the derivation can be a great way to reinforce your understanding and remember it more easily.
7. Use mnemonics: Creating mnemonics or acronyms can be a helpful way to remember a series of steps in a derivation.

Remember, it takes time and consistent effort to remember physics derivations. Keep practicing and seeking help when needed, and you will likely see improvement over time.

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