Question

18. 20 points in the figure below, suppose the current in an infinitely long straight wire is moving towards the left and is given by I = Lot-e-It, where I, and are constants. Everything will be changing slow enough that our static field assumptions are fine. (a) Find the induced current, including direction, in the rectangular loop shown in the figure below if the loop has a resistance of R. (b) Assuming the loop is fixed in space, find the force on the loop as a function of time.

Answer 1

here the direction of B Carring wire is into Paper due to current the plane of = Bds As B is variable to take a Element do form a distance frum wiro anel wente e for that small Element and then Integrate from and to deate dic d = Mora in de

da do Doble de la policy Now Emf = add dt s a mojoa tehtalate) 20 indtc) at 20 3. Hofamine ) E -- Mopoa m (de) [2482€ Atene) here - ve sign show the direction y emf in offoside to clorection of change in flux here as flux decreas with time In will In ward be In clock direction wise so current direction.

VEIR I = V 더 R 's pelo fate ] e14 (2-2) Do the Here the force on segment Ad careteer com celed By BC Now force on AB fi (dexB) F = Px ax Mo I 21(l+C) Now force on co f2 = I'xax Mos ha Net force on trop fonet = fi- fa

(+ Frei - f2- f ( 22> Fi) = 'a thies cance) - sales de ft der er en de *'(234) 35o e PRO Thle. Momoa? t'em*226) en GAR
in order to solve this problem i have used elementry method because magnetic field is variable for loop so i have taken an element and by integrating that element i havr got net magnetic field

after i have use foemula

$\phi =\int B.ds$

and emf=-d$\phi$/dt

and force on current carring conductor is

f=i(dl$\times$B)