- #1
annamal
- 387
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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}##?
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