Question

Shown in the figure below is a long solenoid. Your solenoid has N loops, a length of L, and is carrying a current of I. We shall use the "long" approximation for which the field outside the solenoid is very very small compared to the field inside the solenoid. Use an Ampere path that extends the full length of the solenoid and closes outside the solenoid. N turns in the coil www00000000000) L A "Long" Solenoid (i.e. length >> diameter) What is the length of your Ampere path over which the field is not negligible? What is the current though this path? Use Ampere's Law to determine the FORMULA for the magnetic field inside the solenoid: What is the direction of the magnetic field inside the solenold? Toward the right Toward the left Clockwise as viewed from the left Counter-clockwise as viewed from the left Out of the screen Into the screen Toward the top of the screen Toward the bottom of the screen The solenoid in the picture looks like a "stretched out" spring because the loops of the coil are far apart. If you compressed the coil so that the loops are touching (same number of loops, same current), what would happen to the magnetic field? The field would increase The field would briefly increase then settle to the original value. The field would decrease Not enough information given. The field would stay the same. An EMP would be generated disabling nearby computer equipment. The field would briefly decrease then settle to the original value.

Answer 1

the Q. giver length of solenoid e which very-very large Compare to diomeler (0) ieces>D) The curent 'in solenoid is Number of turns of loobs =N ja IPUIN mm we can show this image by another way B3 l- l (A = carent into R the page lo=0 Bi で 0 o = curent cut of the page corsidey a omperior toob (rectangular amberian loop) of length 'l The magnetic field is alorg Z direction (by Right Hard Rule) tfel The Tength of Amberian li tol, tl, tly from Amberian law = Molenc SB. de + for de for at Ø Bidé 0-0 o=90 0 = 180 0-90 closed line

blo B But So Where ő is magnetic field Tere total canent enclosed by this amperlan loob Os angle and length element so B, li + 0 - 8 12 to fride' By = 0 where Bi= field inside solenoid Bol - Odg = GB. De By = field outside solenoid ( B₂ xo) given Bil = $5.de So The length of Amberian loop for which for which mognetic exist is li=d The cussent through this loop of enclosed in this loob NIM ( cuvent due to N loobs) The total enclosed consent by rect angular loop = Ni field Tere Iene fe Bil q Ni ..so Amberis law de = M Tene B. = M Cla The magnetic field inside the solenoid Bi = Moni where Number of looks her n Bi = Nonio unit length. • To find the disection of ure Right Hand Rute Put thumb of Right Hand along cament and care finger of Right Hards gives in -z axis leftwards os

So The magnetic field is field is along left side of page mg inside is on Bi = Bin a uni con some length still • The mognetic field which depends run (No density of locbs) (cobert) When we compressed the solenold • The eursent (il The Number of turns of look her onit be same i n will be same. os consent enclosed will be same ) so The magnetic field will he same there No charge in magnetic flux through solenoidal across seetion cos os B, is some) so No emf Penseled. The correct option is "The field would stay same. ОХ (os • As is