Electric current and magnetism

3 Likes

@LaveshGupta @AshWin @manan_uppadhyay help bro

2 Likes

Flux nikalke, diff. wrt time. E(ind) aa jayega individual loop ke liye. Initial current ke liye t=0 rakh denge. Rest is current electricity.


Baaki khana khake batata hu bro…

3 Likes

@Neel Whats the answer , i am getting Current in R3 as 9/110 and in R2 as 8/110 A

3 Likes

Do you reverse the current’s side if you get a negative EMF value or just make it positive and continue with right hand sign rule?

3 Likes

@YoDay If the emf is negative then direction of terminals will opposite to the ones which are positive after applying lenz law.

3 Likes

\color{green}{That \;way \;I \;am \;getting}{\frac{1}{110}} \color{yellow}{Amperes\; in\; R3\; then} :frowning:

3 Likes

What was your direction for emf for both the loops ? I mean negative to positive was up side or downside
Even i can be wrong unless Neel confirms the answer.
I think my R2 current might be wrong.

3 Likes

Wait I think I messed up in emf directions. I am trying again now.

3 Likes

Just to confirm, emfs where 0.4V?

3 Likes

Yes but whats your direction coming for both the loops? In the sense its —/- or -/— which means negative to positive is up and down respectively

3 Likes

For the left loop, positive towards the bottom and for the right loop, positive towards above…

3 Likes

Mine positive is coming towards bottom for both.
For first loop B is decreasing and into page and for second B is increasing and out of page so they have same direction no?
I used induced Mag field due to current will oppose change so loop1 mein agar decrease ho raha hain into the page toh induced one will increase it ( into page ) and in second loop field in increasing ( out of page) toh ind mag field will decrease it ( out of page) @YoDay

3 Likes

I just considered out of the page and took positive towards upper side for the loop at the right hand side. I may be wrong in this case.

3 Likes

@AshWin We know that \epsilon=\frac{d(\phi)}{dt} but sometimes we use \epsilon=|\frac{d(\phi)}{dt}| so battery won’t change directions right (in the left loop)? Probably this is wrong in our approach?

3 Likes

@AshWin bro answer is given as R2=13/110 A
But I’ll confirm and let you know

3 Likes

@AshWin bro if possible can you please send image of diagram?

3 Likes

Thanks a lot bro :facepunch:

2 Likes

Thanks bro…but aage ka please batao :sweat_smile:

2 Likes

Bro is it a question from drill test? I came across it one day.

2 Likes