Question

A conductor of length $2.5m$ with one end located at $z=0,x=4m$ carries a current of $12A$ parallel to the negative y-axis. Find the magnetic field in the region if the force on the conductor is $1.2\times {10}^{-2}$ N in the direction $\frac{-\hat{j}+\hat{k}}{\sqrt{2}}$.

Answer 1

Let the uniform magnetic field applied be
$\overrightarrow{B}=B_x\hat{i}+B_y\hat{j}+B_z\hat{k}$
$\Rightarrow d\overrightarrow{F}=i(d\overrightarrow{l}\times \overrightarrow{B})$
$d\overrightarrow{F}=idl(-\hat{j})\times (B_x\hat{i}+B_y\hat{j}=B_z\hat{k})$
$\Rightarrow F_x\hat{i}+F_z\hat{k}=iL{B}_x\left(\hat{K}\right)+iL{B}_z(-\hat{i})$
$\Rightarrow \frac{1.2\times {10}^{-2}}{\sqrt{2}}(-\hat{i}+\frac{1.2\times {10}^{-2}}{\sqrt{2}}(\hat{k})=iL{B}_x(\hat{k})+iL{B}_z(-i)$
$\therefore iL{B}_x=\frac{1.2\times {10}^{-2}}{\sqrt{2}}$
$\therefore B_x=\frac{1.2\times {10}^{-2}}{\sqrt{2}\times 12\times 2.5}T$ and $iL{B}_2=\frac{1.2\times {10}^{-2}}{\sqrt{2}}T$
$\therefore B_z=\frac{1.2\times {10}^{-2}}{\sqrt{2}\times 12\times 2.5}T$
$\therefore \overrightarrow{B}=[\frac{1}{25\sqrt{2}}\times {10}^{-2}(\hat{i})+\frac{1}{25\sqrt{2}}\times {10}^{-2}(\hat{k}\left)\right]T=\frac{{10}^{-2}}{25\sqrt{2}}[\overrightarrow{K}+\overrightarrow{i}]T$;
$B_y$ is arbitary.