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

Answer:

$B = 4.67 \times 10^{-4} T$

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

$i_{en} = 12 - \frac{12}{4^2 - 2^2}(3^2 - 2^2)$

$i_{en} = 7 A$

now from above equation we have

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

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

$B = \frac{\mu_0 7}{2\pi r}$

$B = \frac{(2\times 10^{-7}) 7}{0.003}$

$B = 4.67 \times 10^{-4} T$