# Brilliant Electromagnetism Contest

Hello everyone!

Here's the first ever Brilliant Electromagnetism contest! Hope you have fun and get electrical!

Some of the easy to follow rules of the contest are as follows:

• Questions posted in the contest need not be original.

• The questions could be of any type, starting from proof to MCQs. but you need to post the entire solution for it.

• Scope of the problems is NSEP and JEE Mains/Advanced

• The problem poster needs to know the solution of the problems he poses, and hence needs to post them if no one can answer them within 16 hours of the problem posting.

• One who answers the question posted by me shall post the next question, if his solution is correct.

• One can post the next question in the contest, if the problem poster approves his solution.

And most importantly, have fun!

Best,

Swapnil Das

Note by Swapnil Das
4 years, 4 months ago

This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.

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Invitation to @Mardokay Mosazghi

You are allowed to post the next question, too.

- 4 years, 4 months ago

Problem 3:

Manier times, it is found desirable to have a voltmeter with multiple ranges.

A simple schematic design for the ranges $2V, 20V, 200V$ is shown below.

The galvanometer has resistance $10 \Omega$ and range $0-1 \, mA$

Calculate $R_i \quad i=1,2,3$

- 4 years, 4 months ago

Thanks for maintaining rules and regulations of the contest. I'm inviting certain people to the contest, and if they don't respond, I shall declare the end of the same.

- 4 years, 4 months ago

Since no one has posted a solution in 16 hours I'll post the solution.

Let $V_1 = 2 V \\ V_2 = 20 V \\ V_3 = 200 V \\ R_0 = 10 \Omega \\ i = 1 \, mA$

Then, we get that $i=\frac{V_k}{\sum_{j=0}^{k} R_j } \quad k = 1,2,3$

Solving, we get $R_1 = 1990 \Omega \\ R_2 = 18 k\Omega \\ R_3 = 180 k\Omega$

- 4 years, 4 months ago

Problem 2

The linear charge density on a dielectric ring of radius $R$ varies with $\theta$ as $\lambda=\lambda_0\cos\dfrac{\theta}{2}$ , where $\lambda_0$ is a constant. Calculate the potential at the center $O$ of the ring in volts.

- 4 years, 4 months ago

As total charge on the ring is,

$dq = \lambda_{0} cos(\dfrac{\theta}{2})$

$Q = \int_{0}^{2\pi} \lambda_{0} cos(\dfrac{\theta}{2}) d\theta = 0.$

So potential is also zero at anywhere.

- 4 years, 4 months ago

\begin{aligned} dV &= \frac{1}{4 \pi \epsilon_0} \frac{\lambda_0 \cos \frac{\theta}{2} R\, d\theta}{R} \\ \implies V &= \frac{1}{4 \pi \epsilon_0} \int_{0}^{2 \pi } \lambda_0 \cos \frac{\theta}{2} \, d\theta \\ &= \boxed{0} \end{aligned}

- 4 years, 4 months ago

- 4 years, 4 months ago

@Rajdeep Dhingra @Harsh Shrivastava @Mehul Arora @Deeparaj Bhat and all others are invited to this contest!

- 4 years, 4 months ago

Problem 1

A solid sphere is of uniform charge density and radius $R$.Electric field at some point $P$ inside the sphere is $E$. Now a cavity of sphere is made having centre at point $P$ and radius $r$,$(r. At any point inside this cavity, the electric field is $E_{1}$ and is uniform. So calculate $\dfrac{E}{E_{1}}$.

- 4 years, 4 months ago

$E=$ Electric Field at point P lying inside sphere

Let the sphere be made of infinite uniformly charged shells with radius varying from 0 to R and charges $q_1 , q_2 ,q_3 ,...,q_{R}$ and charge density $\rho$

$E = E_{inside shells} + E_{outside shells}$

= $E_{ inside shells}$

=$\dfrac{Kq_1}{r^{2}}+\dfrac{Kq_2}{r^{2}}+...+\dfrac{Kq_r}{r^{2}}$

=$\dfrac{Kq_{inside}}{r^{2}}$

=$\dfrac{1}{4π\epsilon_0r^{2}} \times \rho \times \dfrac{4πr^{3}}{3}$

=$\dfrac{\rho r }{3\epsilon_0}$

Let the centre of sphere & cavity be $O$ and $P$ respectively. The sphere with cavity is equivalent to system of 2 spheres , one sphere of Radius R & charge density $\rho$ and another sphere of radius $r$ & charge density $-\rho$ coinciding with the cavity.

$E_1 = E_{\rho}+E_{-\rho}$

=$\dfrac{\rho r}{3\epsilon_0}+0$

$\dfrac{E}{E_1}=\boxed{1}$

- 4 years, 4 months ago

thats right i beloeve right @Akash Shukla

- 4 years, 4 months ago

- 4 years, 4 months ago

Hi Aditya! You are requested to post the complete solution so that you can post the next question. Post it whenever you like.

- 4 years, 4 months ago

Yes. you are right

- 4 years, 4 months ago

- 4 years, 4 months ago

No.

- 4 years, 4 months ago

Problem 0:

A stream of protons and deuterons in a vacuum chamber enters a magnetic field. Both protons and deuterons and have been subjected to the same accelerating potential, the kinetic energies of the particles are same. If the ion stream is perpendicular to the magnetic field and the protons move in a circular path of radius ${r}_{p}$, find the radius of the path traversed by the deuterons.

- 4 years, 4 months ago

Let , mass of proton $= m$, mass of deuteron = $2m$, Magnetic field = $B$

Charge of both particles $= q$, velocity of proton = ${v_{p}}$, velocity of deuteron = ${v_{d}}$

As they are accelerated to same potentials and their kinetic energy are same(if we consider the given potential energy is completely transferred to kinetic energy, then they always have same kinetic energy because they both have same charge), so

$m{v_{p}}^2 = 2m{v_{d}}^2$ $\implies$ $v_{p} = \sqrt{2} v_{d}$

The magnetic field is same for both the particles. As they trace the circular path, their centripetal force will be equal to magnetic force which are in opposite direction because the their velocity and magnetic field are both perpendicular.

So, $qvB = \dfrac{m{v}^2}{r}$ $\implies$ $r=\dfrac{mv}{qB}$,

As , $q, B$ are same for both, $r∝ mv$

$∴ \dfrac{r_{p}}{r_{d}} = \dfrac{m{v_{p}}}{2m{v_{d}}}$

$∴ r_{d} = \sqrt{2} r_{p}$

- 4 years, 4 months ago

- 4 years, 4 months ago

is the answer $\sqrt{2}$ ${r}_{p}$ ? @Swapnil Das

- 4 years, 4 months ago

That's what I got as well.

- 4 years, 4 months ago