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*

## Comments

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TopNewestProblem 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 \(R_i \quad i=1,2,3\) – Deeparaj Bhat · 1 year ago

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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 \[ R_1 = 1990 \Omega \\ R_2 = 18 k\Omega \\ R_3 = 180 k\Omega\] – Deeparaj Bhat · 1 year ago

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– Swapnil Das · 1 year 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.Log in to reply

Problem 2The 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. – Aditya Chauhan · 1 year ago

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– Deeparaj Bhat · 1 year ago

\[\begin{align} dV &= \frac{1}{4 \pi \epsilon_0} \frac{\lambda_0 \cos \frac{\theta}{2} R\, d\theta}{R} \\ \implies V &= \frac{1}{4 \pi \epsilon_0} \int_{0}^{2 \pi } \lambda_0 \cos \frac{\theta}{2} \, d\theta \\ &= \boxed{0} \end{align}\]Log in to reply

– Swapnil Das · 1 year ago

Please post the next question.Log in to reply

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

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

So potential is also zero at anywhere. – Akash Shukla · 1 year ago

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Problem 1A 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 \(r\),\((r<R)\). At any point inside this cavity, the electric field is \(E_{1}\) and is uniform. So calculate \(\dfrac{E}{E_{1}}\). – Akash Shukla · 1 year ago

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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.

\(E_1 = E_{\rho}+E_{-\rho} \)

=\(\dfrac{\rho r}{3\epsilon_0}+0\)

\(\dfrac{E}{E_1}=\boxed{1}\) – Aditya Chauhan · 1 year ago

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@Akash Shukla – Mardokay Mosazghi · 1 year ago

thats right i beloeve rightLog in to reply

– Aditya Chauhan · 1 year ago

Is the answer 1?Log in to reply

– Swapnil Das · 1 year ago

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

– Akash Shukla · 1 year ago

Yes. you are rightLog in to reply

– Abhijeet Vats · 1 year ago

Is the answer ((R/r)^3-1)^-1?Log in to reply

– Akash Shukla · 1 year ago

No.Log in to reply

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. – Swapnil Das · 1 year ago

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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, \(r∝ mv\)

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

\(∴ r_{d} = \sqrt{2} r_{p}\) – Akash Shukla · 1 year ago

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– Swapnil Das · 1 year ago

Your answer is correct. Please post the next question.Log in to reply

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@Akash Shukla to post the next question. – Swapnil Das · 1 year ago

Yes, I did askLog in to reply

@Swapnil Das – Shivam Mishra · 1 year ago

is the answer \(\sqrt{2}\) \({r}_{p}\) ?Log in to reply

– Abhijeet Vats · 1 year ago

That's what I got as well.Log in to reply

Invitation to @Mardokay Mosazghi

You are allowed to post the next question, too. – Swapnil Das · 1 year ago

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@Rajdeep Dhingra @Harsh Shrivastava @Mehul Arora @Deeparaj Bhat and all others are invited to this contest! – Swapnil Das · 1 year ago

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