Symmetry – JEE First

Ampere’s law and symmetry

October 31st, 2022jeefirst0 Comments

Problem 3.231 from Irodov.

A current $I$ flows along a lengthy straight wire, as shown in the figure below. From the point O the current spreads radially all over an infinite conducting plane perpendicular to the wire. Find the magnetic field above and below the plane.

Rendered by QuickLaTeX.com — A wire carrying current to an infinite conducting plane. We indicate the radial current on the plane with a few lines, but there is current flowing along every point of the plane.

Solution:

We will use this problem to demonstrate the use of Ampere’s law, starting with some simple scenarios. Our goal is to understand …

Power dissipated in resistance networks

July 8th, 2022jeefirst0 Comments

JEE Advanced 2008 Paper 1, Question 25

The figure below shows three resistor configurations R1, R2 and R3 connected to $3 \mathrm{~V}$ battery. If the power dissipated by the configuration R1, R2 and R3 is P1, P2 and P3 respectively, then

Rendered by QuickLaTeX.com

$P1 > P2 > P3$
$P1 > P3 > P2$
$P2 > P1 > P3$
$P3 > P2 > P1$

Solution

The power dissipated by a resistance network is

(1) $\begin{equation*} P = \frac{V^2}{R} . \end{equation*}$

The potential applied to all three networks arre the same, so we just need to find the resistances. Of circuits R2 and R3, the latter has a greater resistance since

(2) $\begin{equation*} % R_3 = 1 \, \Omega + \frac{2 \, \Omega \cdot 2 \, \Omega}{2 \, \Omega + 2 \, \Omega} = 2 \, \Omega , R_3 = 1 \, \Omega + \left( \frac{1}{2 \Omega} + \frac{1}{2 \Omega} \right)^{-1} = 2 \, \Omega , \end{equation*}$

whereas

(3) $\begin{equation*} % R_2 = \frac{1 \, \Omega \cdot 1 \, \Omega}{1 \, \Omega + 1 \, \Omega} = 0.5 \, \Omega . R_2 = \left( \frac{1}{1 \Omega} + \frac{1}{2 \Omega} + \frac{1}{2 \Omega} \right)^{-1} = 0.5 \, \Omega. \end{equation*}$

The first circuit can be redrawn in the following way to make the comparison easier.…

An infinite ladder of resistors

January 16th, 2022jeefirst0 Comments

Problem 3.152 of Irodov

The figure below shows an inifite circuit formed by the repetition of the same link, consisting of resistance $R_1 = 4.0 \ \Omega$ and $R_2 = 3.0 \ \Omega$ . Find the resistance of this circuit between points $A$ and $B$ .

Rendered by QuickLaTeX.com

Solution

Let’s denote the resistance between the points $A$ and $B$ by $R_{AB}$ . Since the circuit is infinite, removing the first $R_1$ and $R_2$ resistors gives the same arrangement back again — the arrangement is self-similar. That means, the resistance between the points $C$ and $D$ is just $R_{AB}$ without the left-most $R_1$ and $R_2$ resistors, and we may redraw the circuit as shown below.

Rendered by QuickLaTeX.com

It is now straightforward to calculate the resistance,

(1) …

Charge at one corner of a cube

January 8th, 2022jeefirst0 Comments

A charge $q$ sits at one corner of a cube of side length $a$ as shown in the figure below. What is the electric flux through the shaded side?

Rendered by QuickLaTeX.com

Solution

Since the problem is asking us to find the electric flux it is natural to guess that we need to apply Gauss’s Law. However, we need a closed surface with appropriate symmetry to be able to use Gauss’s Law (see article). So we consider the situation shown below.

Rendered by QuickLaTeX.com

We have placed …

Electric field from a sphere with a cavity

January 8th, 2022jeefirst0 Comments

A sphere of radius $a$ is filled with positive charge with uniform density $\rho$ . Then a smaller sphere of radius $a/2$ is carved out, as shown in the figure below, and left empty. What are the direction and magnitude of the electric field at $A$ ? At $B$ ?

Rendered by QuickLaTeX.com

Solution

This problem can be solved by using the principle of superposition. For instance, consider a point charge $+q$ at some point $P$ in space. It creates an electric field everywhere. However, if you place a …