Question

a hollow cylindrical conductor of inner radius 0.0144 m and outer radius 0.0225 m has a magnetic field of magnitude 7.20 × 10 - 5 t at radius of 0.0177 m. what is the current through the conductor? a

Explanation:

Step1: Recall Ampere's Law

Ampere's Law is given by \(\oint\vec{B}\cdot d\vec{l}=\mu_0I_{enc}\). For a point inside the conducting material of a hollow cylindrical conductor (where \(r\) is between the inner radius \(a\) and outer radius \(b\)), the current enclosed \(I_{enc}\) can be found using the formula \(I_{enc}=I\frac{r^{2}-a^{2}}{b^{2}-a^{2}}\). But if we consider a point outside the conductor, \(\oint\vec{B}\cdot d\vec{l}=B(2\pi r)=\mu_0I\), so \(I = \frac{B(2\pi r)}{\mu_0}\). Here, the radius \(r = 0.0177m\), \(B=7.20\times 10^{-5}T\), and \(\mu_0 = 4\pi\times 10^{-7}T\cdot m/A\).

Step2: Substitute values into the formula

Substitute \(B = 7.20\times 10^{-5}T\), \(r=0.0177m\), and \(\mu_0 = 4\pi\times 10^{-7}T\cdot m/A\) into \(I=\frac{B(2\pi r)}{\mu_0}\).

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Answer:

\(6.37A\)