Magnetics!!!

A simple Pendulum with a charged bob is oscillating as shown in figure. Time period of oscillation is $T$ and the angular amplitude is $\theta$ . If a uniform magnetic field perpendicular to the plane of oscillation is switched on, then

$(a)$ $T$ will decrease but $\theta$ will remain constant.

$(b)$ $T$ will remain constant but $\theta$ will decrease.

$(c)$ Both $T$ and $\theta$ will remain the same.

$(d)$ Both $T$ and $\theta$ will decrease.

Easy Math Editor

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Comments

unrelated. but I love your microsoft paint skills

Carlos Spencer - 8 years ago

why unrelated??....but i too appreciate his skill in "painting"....:)....but a suggestion to him::: he could have used "text" option to write the letter "B" in ms. paint.....:-)

Sayan Chaudhuri - 8 years ago

hmm... next tym i will!!! :P

Advitiya Brijesh - 8 years ago

:P LOL!!!

Advitiya Brijesh - 8 years ago

hahaha but I don't think that helped him much!! I think the answer's (e)

Jess J - 8 years ago

LOL!!

Advitiya Brijesh - 8 years ago

If you consider in terms of 'energy consevation', work done by 'magnetic field' is $0$ . Hence, $\theta$ remains same and so or more time also.

A R - 8 years ago

Since, the charged particle is undergoing a periodic oscillation it will be produce only electric field as the displacement is zero after 1 time period!!! Thus, electric field will be opposed by magnetic field by Lenz's law and Lorentz force concept: lets see how.... F=q.E+(-q.vxB) Now see the net force acting on the charged particle is getting reduced by damping force from the magnetic field!!! So, it will oscillate only to balance tension of the string with perpendicular gravitational force acting on it!!! My idea that time period T of oscillation will decrease but \theta\ will remain constant!!! And moreover vXB is maximum on account that B is perpendicular to the plane of oscillation!! So, it is option a)!!!

Subhrodipto Basu Choudhury - 8 years ago

Oscillating charge produces electric field which in turn produces oscillating magnetic field which shoud be true according to you, sir. But one thing to note is that the charges are in 'constantly accelerated' motion, which in this case is equal to $0$ .

Anup Raj - 8 years ago

I guess (c) is correct. Is it right?

Lokesh Sharma - 8 years ago

yes bt how?

Advitiya Brijesh - 8 years ago

Since the restoring force which is mg $\sin$ $\theta$ (' $\theta$ ' as depicted in your diagram) remains the same so will $\omega$ i.e. angular frequency and so will T. Using energy conservation $\theta$ will also remain same since work done by magnetic force is zero.

Nishant Sharma - 8 years ago

$(c)$ is the correct option... I need explanation!!

Advitiya Brijesh - 8 years ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$