# Chapter 16 - Vector Calculus - 16.2 Line Integrals - 16.2 Exercises - Page 1126: 52

$B=\dfrac{\mu_{0} I}{2 \pi r}$

#### Work Step by Step

Here, we have $\int_C B \cdot dr =\int_a^b B r'(t) dt$ Consider $r(t)=r \cos t(t) i+ r\sin (t) j$ $\int_C B \cdot dr =\int_0^{2 \pi} (-B \sin (t) i+ B \cos (t) j) (-r \sin (t) i+ r\cos (t) j) dt$ $= \int_0^{2 \pi} Br ( \sin^2 t+ \cos^2 t) dt$ $= 2 \pi r B$ Ampere's Law states that $\int_C B \cdot dr = 2 \pi r B= \mu_{0} I$ Thus, we have $B=\dfrac{\mu_{0} I}{2 \pi r}$

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