To produce a uniform magnetic field directed | vedclass

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Question

(B) For a long cylinder with uniform current density $J$,the magnetic field at a point $\vec{r}$ from the axis is given by $\vec{B} = \frac{\mu_0}{2}

Vedclass · Accepted Answer

$\frac{\mu_0 J a}{2}$ along $y$-axis

Vedclass · Answer

(B) For a long cylinder with uniform current density $J$,the magnetic field at a point $\vec{r}$ from the axis is given by $\vec{B} = \frac{\mu_0}{2} (\vec{J} 	imes \vec{r})$.Let the two cylinders have current densities $\vec{J}_1 = J \hat{z}$ and $\vec{J}_2 = -J \hat{z}$.Let $\vec{r}_1$ and $\vec{r}_2$ be the position vectors of a point inside the overlap region relative to the axes of the two cylinders respectively.The magnetic field due to the first cylinder is $\vec{B}_1 = \frac{\mu_0}{2} (\vec{J}_1 	imes \vec{r}_1) = \frac{\mu_0 J}{2} (\hat{z} 	imes \vec{r}_1)$.The magnetic field due to the second cylinder is $\vec{B}_2 = \frac{\mu_0}{2} (\vec{J}_2 	imes \vec{r}_2) = \frac{\mu_0 J}{2} (-\hat{z} 	imes \vec{r}_2) = -\frac{\mu_0 J}{2} (\hat{z} 	imes \vec{r}_2)$.The net magnetic field is $\vec{B}_{net} = \vec{B}_1 + \vec{B}_2 = \frac{\mu_0 J}{2} [\hat{z} 	imes (\vec{r}_1 - \vec{r}_2)]$.From the geometry,$\vec{r}_1 - \vec{r}_2 = \vec{a}$,where $\vec{a}$ is the vector from the axis of the second cylinder to the first. Given the orientation in the figure,$\vec{a}$ is along the $x$-axis,so $\vec{a} = a \hat{x}$.Thus,$\vec{B}_{net} = \frac{\mu_0 J}{2} (\hat{z} 	imes a \hat{x}) = \frac{\mu_0 J a}{2} \hat{y}$.Therefore,the magnetic field is $\frac{\mu_0 J a}{2}$ along the $y$-axis.

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