Question

Substituting \[B=\frac{\mu_0 I}{2\pi r}\] into \[\oint\vec{B}\cdot d\vec{l}=B(2\pi r)\] for a straight wire yields:

विकल्प

• \[\mu_0 I\]

• \[\mu_0 I r\]

• \[2\pi\mu_0 I\]

• \[\frac{\mu_0 I}{2}\]

Answer 1

Substituting gives \[\oint\vec{B}\cdot d\vec{l}=\frac{\mu_0 I}{2\pi r}(2\pi r)=\mu_0 I\], exactly as Ampere's Circuital Law predicts.