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*

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## 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 \(R_i \quad i=1,2,3\)

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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 \[ R_1 = 1990 \Omega \\ R_2 = 18 k\Omega \\ R_3 = 180 k\Omega\]

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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.

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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.

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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, \(r∝ mv\)

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

\(∴ r_{d} = \sqrt{2} r_{p}\)

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Your answer is correct. Please post the next question.

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

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That's what I got as well.

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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 \(\dfrac{E}{E_{1}}\).

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\(E=\) 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.

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

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

\(\dfrac{E}{E_1}=\boxed{1}\)

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thats right i beloeve right @Akash Shukla

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Is the answer 1?

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Yes. you are right

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

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Is the answer ((R/r)^3-1)^-1?

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No.

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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.

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

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Please post the next question.

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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.

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

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Invitation to @Mardokay Mosazghi

You are allowed to post the next question, too.

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