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
3 years, 2 months ago

No vote yet
1 vote

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Problem 3:

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

A simple schematic design for the ranges 2V,20V,200V2V, 20V, 200V is shown below.

The galvanometer has resistance 10Ω10 \Omega and range 01mA0-1 \, mA

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

Deeparaj Bhat - 3 years, 2 months ago

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

Let V1=2VV2=20VV3=200VR0=10Ωi=1mAV_1 = 2 V \\ V_2 = 20 V \\ V_3 = 200 V \\ R_0 = 10 \Omega \\ i = 1 \, mA

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

Solving, we get R1=1990ΩR2=18kΩR3=180kΩ R_1 = 1990 \Omega \\ R_2 = 18 k\Omega \\ R_3 = 180 k\Omega

Deeparaj Bhat - 3 years, 2 months ago

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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.

Swapnil Das - 3 years, 2 months ago

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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 rp{r}_{p}, find the radius of the path traversed by the deuterons.

Swapnil Das - 3 years, 2 months ago

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Let , mass of proton =m= m, mass of deuteron = 2m2m, Magnetic field = BB

Charge of both particles =q= q, velocity of proton = vp{v_{p}}, velocity of deuteron = vd{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

mvp2=2mvd2m{v_{p}}^2 = 2m{v_{d}}^2     \implies vp=2vdv_{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=mv2rqvB = \dfrac{m{v}^2}{r}     \implies r=mvqBr=\dfrac{mv}{qB},

As , q,Bq, B are same for both, rmvr∝ mv

rprd=mvp2mvd∴ \dfrac{r_{p}}{r_{d}} = \dfrac{m{v_{p}}}{2m{v_{d}}}

rd=2rp∴ r_{d} = \sqrt{2} r_{p}

Akash Shukla - 3 years, 2 months ago

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Your answer is correct. Please post the next question.

Swapnil Das - 3 years, 2 months ago

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

shivam mishra - 3 years, 2 months ago

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That's what I got as well.

Abhijeet Vats - 3 years, 2 months ago

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Problem 1

A solid sphere is of uniform charge density and radius RR.Electric field at some point PP inside the sphere is EE. Now a cavity of sphere is made having centre at point PP and radius rr,(r<R)(r<R). At any point inside this cavity, the electric field is E1E_{1} and is uniform. So calculate EE1\dfrac{E}{E_{1}}.

Akash Shukla - 3 years, 2 months ago

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E=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 q1,q2,q3,...,qRq_1 , q_2 ,q_3 ,...,q_{R} and charge density ρ\rho

E=Einsideshells+EoutsideshellsE = E_{inside shells} + E_{outside shells}

= EinsideshellsE_{ inside shells}

=Kq1r2+Kq2r2+...+Kqrr2\dfrac{Kq_1}{r^{2}}+\dfrac{Kq_2}{r^{2}}+...+\dfrac{Kq_r}{r^{2}}

=Kqinsider2\dfrac{Kq_{inside}}{r^{2}}

=14πϵ0r2×ρ×4πr33\dfrac{1}{4π\epsilon_0r^{2}} \times \rho \times \dfrac{4πr^{3}}{3}

=ρr3ϵ0\dfrac{\rho r }{3\epsilon_0}

Let the centre of sphere & cavity be OO and PP respectively. The sphere with cavity is equivalent to system of 2 spheres , one sphere of Radius R & charge density ρ\rho and another sphere of radius rr & charge density ρ-\rho coinciding with the cavity.

E1=Eρ+EρE_1 = E_{\rho}+E_{-\rho}

=ρr3ϵ0+0\dfrac{\rho r}{3\epsilon_0}+0

EE1=1\dfrac{E}{E_1}=\boxed{1}

Aditya Chauhan - 3 years, 2 months ago

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thats right i beloeve right @Akash Shukla

Mardokay Mosazghi - 3 years, 2 months ago

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Is the answer 1?

Aditya Chauhan - 3 years, 2 months ago

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Yes. you are right

Akash Shukla - 3 years, 2 months ago

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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.

Swapnil Das - 3 years, 2 months ago

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

Abhijeet Vats - 3 years, 2 months ago

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No.

Akash Shukla - 3 years, 2 months ago

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Problem 2

The linear charge density on a dielectric ring of radius RR varies with θ\theta as λ=λ0cosθ2\lambda=\lambda_0\cos\dfrac{\theta}{2} , where λ0\lambda_0 is a constant. Calculate the potential at the center OO of the ring in volts.

Aditya Chauhan - 3 years, 2 months ago

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dV=14πϵ0λ0cosθ2RdθR    V=14πϵ002πλ0cosθ2dθ=0\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}

Deeparaj Bhat - 3 years, 2 months ago

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Please post the next question.

Swapnil Das - 3 years, 2 months ago

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As total charge on the ring is,

dq=λ0cos(θ2)dq = \lambda_{0} cos(\dfrac{\theta}{2})

Q=02πλ0cos(θ2)dθ=0.Q = \int_{0}^{2\pi} \lambda_{0} cos(\dfrac{\theta}{2}) d\theta = 0.

So potential is also zero at anywhere.

Akash Shukla - 3 years, 2 months ago

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

Swapnil Das - 3 years, 2 months ago

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Invitation to @Mardokay Mosazghi

You are allowed to post the next question, too.

Swapnil Das - 3 years, 2 months ago

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