Question

A long, hollow, cylindrical conductor (inner radius = 2.0 mm, outer radius = 4.0 mm) carries a current of 12 A distributed uniformly across its cross section. A long wire which is coaxial with the cylinder carries an equal current in the same direction. What is the magnitude of the magnetic field 3.0 mm from the axis?

Answer 1

`B = 4.67 * 10^(-4) T`

Step-by-step explanation:

As we know that by ampere's law we have

`\int B.dl = \mu_0 i_(en)`

here we know that enclosed current here at radius r = 3 mm is given as

`i_(en) = 12 - (12)/(4^2 - 2^2)(3^2 - 2^2)`

`i_(en) = 7 A`

now from above equation we have

`\int B.dl = \mu_0 (7 A)`

`B(2\pi r) = \mu_0 7`

`B = (\mu_0 7)/(2\pi r)`

`B = ((2* 10^(-7)) 7)/(0.003)`

`B = 4.67 * 10^(-4) T`