Magnetism Doubt

Here is my attempt. Ask anything. I am not sure. :)

#ElectricityAndMagnetism

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Comments

The final flux you calculated is not correct. But you are not on the wrong track. I will share my solution after some time. This is a very nice conceptual question. I do not know if my answer is correct though. You should think about this problem too.

Karan Chatrath - 8 months, 2 weeks ago

@Karan Chatrath Please share your solution . Thanks in advance.
In my opinion your answer is correct. Because my friend was practicing physics for JEE exam.

And in this exam answer is probably a good number like (1, 5,10,12,) it is not like answer will come 3.873,4.273,9.282.

Talulah Riley - 8 months, 2 weeks ago

@Karan Chatrath But How my $\phi_{final}$ is incorrect?

Talulah Riley - 8 months, 2 weeks ago

@Karan Chatrath sorry for the delay.posting attempt till 11:30

Talulah Riley - 8 months, 2 weeks ago

@Karan Chatrath I have uploaded my attempt above.

Talulah Riley - 8 months, 2 weeks ago

@Steven Chase Hello,in case you want to contribute something. You are welcome.

Talulah Riley - 8 months, 2 weeks ago

Say the area vector points into the plane, in the direction of the magnetic field. Consider the initial flux:

$\Phi_{i} = \vec{B} \cdot \vec{A} = Bab$

Now consider the geometry in the final configuration. In the final configuration, had the area of the loop is turned by 180 degrees. The area of one half is:

$\vec{A}_1 = \frac{ab}{4} \hat{k}$ $\vec{A}_2 = -\frac{ab}{4} \hat{k}$

Now, the final flux is:

$\Phi_{f} = \vec{B} \cdot \vec{A}_1 + \vec{B} \cdot \vec{A}_2= 0$

Now, let us apply KVL in the loop. the equation would be:

$\lvert \frac{d\Phi}{dt}\rvert = L\frac{dI}{dt} + IR$

Now, consider the system to move from it's initial state to its final state in a time $\Delta t$ . The initial and final current must be zero. So KVL can also be written as:

$\lvert \frac{\Delta \Phi}{\Delta t}\rvert = L\frac{\Delta I}{\Delta t} + R\frac{\Delta Q}{\Delta t}$

Now, between initial and final states, $\Delta I=0$ , so therefore:

$\lvert \frac{\Delta \Phi}{\Delta t}\rvert = 0+ R\frac{\Delta Q}{\Delta t}$

$\implies \Delta \Phi = R \Delta q$

Karan Chatrath - 8 months, 2 weeks ago

@Karan Chatrath in the 4th step, there are two $=$ .

Talulah Riley - 8 months, 2 weeks ago

Edited solution.

Karan Chatrath - 8 months, 2 weeks ago

@Karan Chatrath – @Karan Chatrath the total final flux will cancel out isn't. Because area vectors are opposite. I am bit confused about that?

Talulah Riley - 8 months, 2 weeks ago

@Talulah Riley – Yes, that is what will happen, according to me.

Karan Chatrath - 8 months, 2 weeks ago

@Karan Chatrath – @Karan Chatrath Can be both area vectors in same. How you are deciding direction?.

Talulah Riley - 8 months, 2 weeks ago

@Talulah Riley – I do not understand your question.

Karan Chatrath - 8 months, 2 weeks ago

@Karan Chatrath – @Karan Chatrath the question is “How you know that the direction of flux through both the triangles are opposite? ”

Talulah Riley - 8 months, 2 weeks ago

@Talulah Riley – This is now I have understood the problem. One of the conducting rods is turned by 180 degrees to achieve the final configuration. So a part of the area is still pointing in its initial direction, but the other half turns by 180 degrees thereby making it oppositely directed.

Karan Chatrath - 8 months, 2 weeks ago

@Karan Chatrath – @Karan Chatrath Oh wow! I was also thinking the little bit same. Thanks for clarification.

Talulah Riley - 8 months, 2 weeks ago

@Karan Chatrath How do you get 5 only?

Talulah Riley - 8 months, 2 weeks ago

@Karan Chatrath I always consider SI units in Physics

Talulah Riley - 8 months, 2 weeks ago

@Karan Chatrath wait one last doubt on this question.
I didn't understand 6th to 7th step.? Why the initial and final currents are 0?

Talulah Riley - 8 months, 2 weeks ago

Because the emf is only induced as long as the loop is turning. Once the loop stops turning, there is no induced emf because flux is no longer changing, so the current decays to zero.

Karan Chatrath - 8 months, 2 weeks ago

@Karan Chatrath Oh yes.Thank you so much.

Talulah Riley - 8 months, 2 weeks ago

@Karan Chatrath I didn't understand this Problem .
I want to know how to find compression and extension.HELP ME WHENEVER YOU WILL BE FREE.
thanks in advance.

Talulah Riley - 8 months, 2 weeks ago

I think I will post a note or a problem related to this problem, later.

Karan Chatrath - 8 months, 2 weeks ago

@Karan Chatrath Thank. I am waiting. By the way my SHM is bit weak. After solving many many problems , I always gets stuck in a new problem.

Talulah Riley - 8 months, 2 weeks ago

@Talulah Riley – The problem with SHM problems is that while preparing for JEE type exams, people use a lot of shortcut methods to solve them. These shortcuts are different for different types of questions and are not very easy to understand. Treat any SHM problem as a regular mechanics problem.

Karan Chatrath - 8 months, 2 weeks ago

@Karan Chatrath – @Karan Chatrath Exactly. You are absolutely correct.
There are many shortcuts in physics which we use in JEE and when we post those problems in Brilliant., no one use that short cut , You and Steven sir always solve those questions from basic method like a physicist.
By the way do you have given JEE exam?

Talulah Riley - 8 months, 2 weeks ago

@Talulah Riley – Yes, I appeared for that exam but I did not clear it. I did not do well.

Karan Chatrath - 8 months, 2 weeks ago

@Karan Chatrath Here is the answer to your all questions respectively. 1)Yes
2)No
3)Would you like to see my solution?

Talulah Riley - 8 months, 1 week ago

I would like to see your approach. No need to post the entire solution.

Karan Chatrath - 8 months, 1 week ago

@Karan Chatrath my approach is only use of kinematics and simple harmonic motion.

Talulah Riley - 8 months, 1 week ago

@Talulah Riley – I want you to describe your approach and not give me a generic answer. Describe your steps.

I analysed the time taken for the two dumbbells to first come into contact.
At that instant, I applied momentum and energy conservation to calculate the final velocities of the two masses that collide, after collision. The other two masses' velocities remain unchanged.
I use the positions and velocities of all the masses as initial conditions to solve the differential equations of motion after collision.

Karan Chatrath - 8 months, 1 week ago

@Karan Chatrath – @Karan Chatrath here are some steps which i hae followed.
1)Momentum conservation.
2)Reduced mass system.
3)Simple harmonic
4)Kinematics.

Talulah Riley - 8 months, 1 week ago

@Talulah Riley – You are not elaborating. Nevermind. I have changed my mind. You may share your solution.

Karan Chatrath - 8 months, 1 week ago

@Karan Chatrath – @Karan Chatrath Sorry I don't think I can elaborate more. Because there is nothing in the question which needs to be elaborated. And my english is also not so good. I will upload the solution on 4:00PM today. Bye

Talulah Riley - 8 months, 1 week ago

@Talulah Riley Hey are you there?

Krishna Karthik - 7 months, 2 weeks ago

@Karan Chatrath I am getting answer as $5×10^{-6}$

Talulah Riley - 8 months, 2 weeks ago

Read the question. The answered is required in $\mu C$ .

Karan Chatrath - 8 months, 2 weeks ago

@Karan Chatrath Sorry, I understand everything. $THANKS$

Talulah Riley - 8 months, 2 weeks 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}`	$\boxed{123}$