Ready for some different action??
Give me the value of the quantity mentioned in the diagram...
i.e the line integral of the magnetic feild (due to the two coaxoial loops) along the common axis of the two current carrying loops.
Assume the radius of both the loops to be some 'R';
Represent the permeability by U*
is the ans 3u* ...... just a guess
ReplyDeleteyes it is 3U*
ReplyDeleteactually it is U*( I1 + I2 ) where I1 and I2 are the currents
You both are right.
ReplyDeleteActually, this question has a fantastic solution..i guess Kapil has applied that?
sorry fr bein late(kal ts hai na!)..yeah im also getting 3u*..but i dint understand fantastic..matlab choosing the correct loop for it is the fantastic thing,?
ReplyDeleteya fantastic is quite relative.. you'll understand it in tomorrows origins.
ReplyDeletehmm. i didn find much strength in the question. no concept, weak integration... lukin forward to the fantastic part though..
ReplyDeletehi vibhav.. plz read digvijay's comment!
ReplyDeletealso i first solved it using integration.
ReplyDeletebut vibhav...would the integration remain as easy if i replaced the 2 loops by a tight solenoid of finite length..N turns..current i??
Anyone wanna have a shot with integration?? (i guess not?)
Not really. i wouldn want to go too deep into integration because wont have that much time in my hand during the jee exam... thats y i logged on in the first place, to find a shorter solution to most things and because its interesting to know the variety of solutions to a problem.
ReplyDeleteyou're right.. some questions are meant to be left in the exam.. and pondered over at home later.
ReplyDeleteBut now..seeing the solution, maybe you could solve the solenoid problem??
wouldn it be the same except u have an extra term N in the answer??
ReplyDeletei mean, u can take a solenoid of N loops with current i to be N rings of current i each...
yup.. so it was better than u first thought?
ReplyDeleteyeah.
ReplyDeletebut something struck me, even wen i solved it the first time.
u havent given the distance between the two rings, which means it doesnt reaally matter in the context of the question.
which implies that the answer would remain the same no matter how close u bring the rings.
and if both the rings had the same current, u can connect the rings as it would hardly make a difference, and u got urself a solenoid!
guess u had to integrate for one ring in the end after all!
but i do agree, ur solution was definitely better.
yaar in the original question, the distance was also given..and i fell for that!
ReplyDelete