Recent content by cheapstrike

Homework Statement The velocity of efflux of a liquid through an orifice in the bottom of the tank does not depend upon: A. Size of orifice B. Height of the liquid C. Acceleration due to gravity D. Density of the liquid Homework EquationsThe Attempt at a Solution My textbook says that...

Yes, into :D

Wait.. lol I also assumed the loop to be in the plane of paper with its axis perpendicular to the plane of paper and B parallel to the axis of the loop. Like generally we draw a loop on paper and take B inside it's plane (x).

Yeah, I mixed up between flexible and elastic. I at first thought that the loop will go on increasing in radius, causing change in flux, till the net current becomes zero.

There is no point in writing this but... I think, saying A and C are correct is fine coz these two are two different cases. In A) only I is being doubled and in C) only r is being doubled. That's why option D) is there which shows that if the asker of the question wants to change two parameters...

I am not sure, i don't even know the answer. But the method seems correct.. :D Thanks a lot

I am a little unsure about this but, I wrote this as dFcosφ=Tθ (i wrote sinθ=θ) and then wrote dF as IdlB where dl is rdφ and integrated it from -θ/2 to θ/2 and wrote sinθ/2=θ/2 and finally got IBr=T. Is this correct? If this is, then shouldn't the answer be A) and C)?

Ok.. so I took the angle that T vector makes incorrectly. It should be dx/2.

I think the force acting radially outwards on the element dl will be I(dlxB) which is IdlB as dl is perpendicular to B. This force will be balanced by sin of both the tension forces. Also, their cosines will cancel each other. So we will have an equation IdlB=2Tsin(dx), where T is the tensional...

Ah.. I got mixed up with that. Ok. So the element dl will remain at it's place meaning net force on it is zero? So, the net outward force = I(dlxB) will be balanced by the two forces at it's ends. Btw how will we do it?

Ok.. so in the FBD, there will be an outward force and a force on both sides of dl due to the other two ends. But can we get any relation from here? I mean the loop is elastic so won't it expand?

Will the force acting on the dl element not be radially outwards?

I think F=I(lxB) where l will be 2*pi*r?

Homework Statement A small circular flexible loop of wire of radius R carries a current I. It is placed in a uniform magnetic field B. The tension in the loop will be double if A) I is doubled B) B is halved C) r is doubled D) both B and I are doubled. Homework Equations The Attempt at a...

Thanks. But I still can't figure how to apply this for the rectangular strip sorry.