- #1

annamal

- 335

- 28

The magnetic flux ##\phi_m = \int{BdA}##

The magnetic field of the coaxial cable B = ##\frac{I_{enc} \mu_0}{2\pi r}##

since surface area of a cylinder = ##2\pi rdr L, dA = 2\pi L dr## where L is the length of the coaxial cable

so ##\phi_m = \int{\frac{I_{enc} \mu_0}{2\pi r}2\pi L dr}##?

The magnetic field of the coaxial cable B = ##\frac{I_{enc} \mu_0}{2\pi r}##

since surface area of a cylinder = ##2\pi rdr L, dA = 2\pi L dr## where L is the length of the coaxial cable

so ##\phi_m = \int{\frac{I_{enc} \mu_0}{2\pi r}2\pi L dr}##?

Last edited: