Simple Harmonic Current

Electricity and Magnetism Level 4

$P$ is a point at a distance of $d= \SI[per-mode=symbol]{1}{\meter}$ from the center $O$ of an earthed conducting sphere of radius $R= \SI[per-mode=symbol]{0.5}{\meter},$ and a charge of $q= \SI[per-mode=symbol]{2}{\coulomb}$ is undergoing simple harmonic motion (SHM) about $P$ along $OP.$

If $A = \SI[per-mode=symbol]{10^{-3}}{\meter}$ is the amplitude of the oscillations, and $\omega= \SI[per-mode=symbol]{5}{\radian\per\second}$ is the angular frequency of the SHM, find the maximum value of current $I$ (in amperes) flowing in the wire $CD$ . Submit your answer to 3 decimal places.

The answer is 0.005.

1 solution

Spandan Senapati
May 30, 2017

Relevant wiki: Conductors

Let the charge $q$ be at a distance $x$ from the center of the sphere. The sphere is earthed, thus it will exchange the charge from the Earth to keep its potential zero. Let the sphere gets a charge $q$ .

Now, the potential of the sphere is zero, thus $\frac{kq}{x} + \frac{kQ}{R} = 0$ $Q=- \frac{qR}{x}.$

The current can be calculated by differentiating the charge with respect to time. $i = \frac{dQ}{dt}$

Hence, the current $I= \dfrac{qR}{x^2}v$ .

Since $A>>d, \, x$ can be treated almost constant now and current will be maximum when the speed of the particle is maximum. We know that the maximum speed of a particle in SHM is $v_{max} = A \omega$ .

So, $I_{max} = \dfrac{qR}{d^2} A \omega$ . Substituting the data in the equation we get the maximum current $\boxed{0.005 \text{ A}}$

Good did the same way.

It was asked in our 'Monday' test.

Harsh Shrivastava - 4 years ago

Hey try this one "supermassive black holes". I posted it some days back.This was a q in APHO 2017

Spandan Senapati - 4 years ago

Bro I don't have sufficient knowledge and skills to solve it I guess, btw I already saw your problem.

Harsh Shrivastava - 4 years ago

@Harsh Shrivastava – Oh you don't need to... The only thing that is needed is gravitation and some energy related to it.....The gravitational power is already given.....And you know this was a q in inpho-2017(it was a star-blackhole system).....I hadn't encountered....But believe me give it a try and you can solve it easily.....

Spandan Senapati - 4 years ago

@Spandan Senapati – I get the following idea:

Use dE/dt and then use Kepler's law to relate w and radius at any instant and then we are done.

Am I correct?

Harsh Shrivastava - 4 years ago

@Harsh Shrivastava – Ya done. That's what the question asked.

Spandan Senapati - 4 years ago

@Spandan Senapati – We will integrate then between required limits.

Harsh Shrivastava - 4 years ago

@Harsh Shrivastava – Can you help me with limits and what is that schwarzradius?

Harsh Shrivastava - 4 years ago

@Spandan Senapati – We can write E = -GM^2/(2a) at any instant.

Harsh Shrivastava - 4 years ago

@Harsh Shrivastava – Give it a try...if not I will give some hints.....

Spandan Senapati - 4 years ago

Simple Harmonic Current

The answer is 0.005.

1 solution

0 pending reports