Along, hollow, cylindrical shell has a current flowing uniformly down its length. the shell has outer radius 4 mm. at a point 3 mm from the axis of the cylindrical shell, the magnetic field strength is measured to be 5/9 of the magnetic field strength at the outer radius. find the inner radius of the shell.

inner radius is 2 mm

Explanation:

given data

outer radius b = 4 mm

x point radius r = 3 mm

to find out

inner radius

solution

we consider here a is inner radius and b is outer radius and x is point of radius 3 mm

we know by given data

B(3mm) = 5/9 B(4mm)     1

so here 3 case when magnetic field at radius less than a and between a and b and last radius = B

so in 1st case

magnetic field = 0 at radius less than a   2

in 2nd case

magnetic field is between a and b

so B = ( µ I / 2πr ) × ( r² - a² / b² - a² )       3

and

in 3rd case

magnetic field is at radius = b

so here B  =  ( µ I / 2πb )                       4

so from equation 1 and 3 , 4

we get

( µ I / 2πr ) × ( r² - a² / b² - a² )  = 5/9  ×  ( µ I / 2πb )  

put here all value and find a

( µ I / 2πr ) × ( 3² - a² / 4² - a² )  = 5/9  ×  ( µ I / 2π4 )  

( 9 -a²)  / (48 - 3a²)  =  5 / 36

solve it and we get

a = 2

so inner radius is 2 mm