A long straight wire carries a current i a rectangular loop...

Name: a long straight wire carries a current i a rectangular loop of side lengths a and b is placed nearby to the wire a distance of d away such that the side of length a runs parallel to the long 14922
Uploaded: 2021-06-18T06:44:13-08:00
Duration: 11 min 17 s
Channel: Vishal Gupta
Description: a long straight wire carries a current i a rectangular loop of side lengths a and b is placed nearby to the wire a distance of d away such that the side of length a runs parallel to the long 14922

A long, straight wire carries a current $I$. A rectangular loop of side lengths $a$ and $b$ is placed nearby to the wire a distance of $d$ away such that the side of length $a$ runs parallel to the long, straight wire. Problems: Part a) Write an expression for the magnetic flux through the rectangular loop. Part b) Assuming the current varies by $2A$ per second, and the rectangular loop has sides $a = 2cm$ and $b = 5cm$ at a distance of $d = 1cm$ away, what is the induced emf in the loop?

Transcript

00:01 In the given problem, we have been given a straight wire carrying a current i in it from left to right and a rectangular loop having width and length l.

00:28 The upper arm of the loop is at a distance edge from the wire.

00:36 In the first part of the problem we have to find the two magnetic flux passing through this loop rectangular loop as the magnetic field due to this current carrying conductor will keep changing the magnetic field depends on the distance and the distance of various segments of this loop from the wire are changing so the flux will keep changing so to find the total flux we will use calculus method means differentiation and integration so first of all we consider a small strip of this rectangular loop whose length is same as l but width is dx and it is at a distance of x from the upper side of the loop so its distance from the wire will become h plus x area small area of this strip will become width dx and length l means l dx so we know the distance so at this distance from the wire the magnetic field will be given as the expression for magnetic field at a distance r from a straight current carrying conductor is given as new not upon four pi 2 i by r so here it will become we knot by 2 pi i by h plus x so the small flux we will consider this magnetic field as constant which will be passing through this segment as the distance d x is negligibly small so there will be no change in the magnetic field passing through this small strip hence this small flux will become b into d a means mu not by 2 pi into i by h plus x and for area l d x so if we want to find the total flux we will integrate the so total magnetic flux will be given as integration of this mu not upon 2 pi i l d x x plus x here the constants will come out so it will become mu not i l by 2 pi and the remaining will be 1 by h plus x d x so as we have to integrate this x and the x is varying from 0 to the maximum w so the limits of this integration will be from 0 to w now the integration of 1 upon h plus x with respect to x comes out to be log base e h plus x limits 0 to w and when i put these limits i'll get log base e h plus w minus log h.

04:53 Means when i put the upper limit w the x will be written as w and in the lower limit 0 x will be written as 0 so it will remain h only log base e so finally the expression for the flux becomes mu not i l by 2 pi log base e means a natural algorithm h plus w by h so this is the answer for the first part of the problem now in the second part we have to find the emf induced through this coil which according to faradhi's laws of electromagnetic induction is given as minus d5 by d t so if i look only for the magnitude it will become the differentiation will be of i only so it will become new not l constant to pi constant log base e h plus w by h constant and differentiation of i which will become d i by d t now the expression for the current is given as a plus pt so here d i by d t will become zero plus b means just b so emf induced will be mu not l by 2 pi log base e into b now we will put all the values so here it will be in place of this mu not upon 2 pi i can put 2 into 10 dash to 4 minus 7 into length 10 1 .00 meter length is 1 .00 meter then logarithm of 1 .00 plus 10 upon 1 .00 and in place of b this is that these are in centimeters so i'm not changing the unit because centimeter will be cancelled by the centimeter so finally i get 2 into 10 dash to minus 7 into 10 log base e 11 and if i use sign convention also everywhere it will be minus so the final answer is minus 2 into 10 power minus 7 into 10 into 2 .3979 volt.

08:58 So finally, this is minus 4 .8 into 10 dashed per minus 6 volt, the induced dmf in the coil...

Question

Please give Ace some feedback