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the current running in the element which is enclosed ba ur imaginary amperian loop......
theoutside current is usually taken positive and inside negative,,,

fantastic n phodu........

gud wrk both.......

yaar i feel sm1 else has also posted it somewhere on goiit.......

me too wanna play it but yaar since i belong to a place now where there is a lot of electricity probs
so ny solution

see:

let

tanx = t²

now:

sec²x dx = 2t dt

dx = 2t dt

-------------------

1 + t^4

and integral

I = t*dx

use dx frm above and proceed....

now it becomes very simple......

it is really purely ur convention which wont affect ur final ans....
now u can either take outside current to be positive and inside one negative,
but let me tell u one thing.....
let u have assigned cw direction 4 ur amperian loop...
then curl ur fingers in the assumed direction and the direction of thumb will give u ur positive direction

for which group do u want????
nudge me???

let one spoon of water be x% volume of the glass of water.
now
when one spoon of water is added to glass of milk;
milk glasss contains one glass of milk + water.
now x% of it will be more in seocond glass.
so i will go with second case as ma answer.

let one spoon of water be x% volume of the glass of water.
now
when one spoon of water is added to glass of milk;
milk glasss contains one glass of milk + water.
now x% of it will be more in seocond glass.
so i will go with second case as ma answer.