Electricity and Magnetism

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 …

Biot-Savart law on a polygon

September 30th, 2022jeefirst0 Comments

A current $I$ flows along a thin wire, shaped as a regular polygon with $N$ sides, which can be inscribed intro a circle of radius $R$ . Find the magnetic field at the center of the polygon. What happens if $N$ is made very large?

Solution:

The polygon described in the problem is illustrated in the figure below, for $N=8$ . The magnetic field created by this structure is the superposition of fields from $N$ straight current carrying wires of length $2 R \sin (\pi/N)$ . Therefore, we will need to find the magnetic field due to a wire of finite length first.

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The Biot-Savart law gives the magnetic field …

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

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

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.…

Electric flux through a hemisphere

March 24th, 2022jeefirst0 Comments

JEE Advanced 2017 Paper 2, Question 10

A point charge $+Q$ is placed just outside an imaginary hemispherical surface of radius ${\rm R}$ as shown in the figure. Which of the following statements is/are correct?

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The electric flux passing through the curved surface of the hemisphere is $-\frac{Q}{2 \varepsilon_{0}}\left(1-\frac{1}{\sqrt{2}}\right)$
Total flux through the curved and the flat surfaces is $\frac{Q}{\varepsilon_{0}}$
The component of the electric field normal to the flat surface is constant over the surface
The circumference of the flat surface is an equipotential

Solution

In problems like these a beginner may …

Resistor network with mirror symmetry

February 26th, 2022jeefirst0 Comments

JEE Advanced 2012 Paper 1, Question 12

For the resistance network shown in the figure, choose the correct option(s).

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The current through $PQ$ is zero.
$I_1 = 3$ A.
The potential at $S$ is less than that at $Q$ .
$I_2 = 2$ A.

Solution

We can speed up the solution by noticing the mirror symmetry of the circuit. Let the current $I_1$ split into two parts $I_a$ and $I_b$ at the node $X$ (see figure). Similarly, let the currents going into node $Y$ be $I_a'$ and $I_b'$ . If we were to reverse the polarity of the $12 \, {\rm V}$ battery connected to the circuit, all currents would simply reverse direction, but their magnitudes would remain the …