Question

Magnetostatic field of on infinite, quasi filiform colinder with longitudinal cavity offe axir. This is a mathematically easy, but montrivial concentual problem. (Hint: Use the superposition principle fore I Consider an infinite, quasi filiforan colinduic conductor with radius a. A stationary current I lower along the longitudinal axis of the conductor, "Che current, which is directed upward, is uniformly distributed through the conductor con- section. Os shown in Gring HA5.9, a colindrical cavity with radiur ala ir gren longitudinally, throughout the entire lenght of the conductor. The cavity is off as is such that its center is the mid point of a radiour of the original conductor, after opening the longitudinal cavity, the initial current density associated with I ir maintained constant.

Answer 1

7on IS Suppose then by definition of 3 we have Q = J Aa² a * J = I Ta² current density F flows throughout the cylinder, I=f7.15 --0 cavity flowing IZ = 42-43 Now a is as shown, I) Now current (1 is is made into the cylinder in portion Je hebt I2 = 5 - 4 then current flowing in pation ② is, I = I - 7 = 32 2 -3 = Jña? 4 2) 3 3 Now we will use superposition principle to calculate magnetic field at different location, - (a) at point o Since no current is enclosed , is field due to whole magnetic cylinder zero, Ience I - 4 Now, magnetic field due to removed portion is lead = Mo Jence (Ampere's cow). Quar च B = Mo I a B = Mo I una 4 (204) Net magnetic field will be > B = - Mo I 4 na in ģ direction. removed b) At point le magnetic field due to of removed portion , so Portion liblat cylinder will be zero as it is center net magnetic field will be due to the whole Holence a Bega B(27) = 40 J Tra? CS Scanned with CamScanner a B = Moja a B = Mo I a a B = Mo I ta ra Som YEYTT

Now at point P magnetic field due to whole portion , 88. d = Mo Ienie ? B,= Mo I 2nd & magnetic field due to removed portion, B.2016–9) Bidt = Molence & Mo I al Bi Ho I 8*(1-2) Bonet 8,-B> Mol Hoz (het -d) złd Bad (d-a/) Mo I 8-10-> Mol. (32-20 Bad (d-4) q 다 B Mo I (3d-za) Brd[d-9) all field are in e direction ( sign incidided) CS Scanned with CamScanner