Question

A short current element d$\overrightarrow{\mathbf{l}}$= 10.500 mm2ne carries a current of 5.40 A in the same direction as d$\overrightarrow{\mathbf{l}}$. Point P is located at$\overrightarrow{\mathbf{r}}$= (-0.730 m)$\hat{\mathbf{i}}$+ (0.390 m)$\hat{\mathbf{j}}$. Use unit vectors to express the magnetic field at P produced by this current element.

Short answer

Using the unit vectors to express the magnetic field at the produced current element is

${\mathrm{dB}}_{\mathrm{x}},{\mathrm{dB}}_{\mathrm{y}},{\mathrm{dB}}_{\mathrm{z}}=\left(1.85.{10}^{-10},0,3.47.{10}^{-10}\right)\mathrm{T}$

Step-by-step solution

Unit vector

Using the Biot-Savart law

$\frac{{\mu }_0}{4\pi }\frac{id\overrightarrow{l}\times \hat{r}}{r^2}$

Now turning into unit vector, we get:

$\frac{\left(-0.73,0,0.39\right)}{\sqrt{\left(0.{73}^2+0.39\right)}}$

Magnetic field at point p

Now after plugging the values in the equation, we get:

$\frac{{\mu }_0}{4\pi }\frac{5.4\left(0,0.0005,0\right)\times \left(-0.88,0,0.47\right)}{0.{83}^2}$

Now computing the cross product by finding the determinant of the matrix we get:

$\begin{array}{ccc}i& j& k\\ 0& 0.0005& 0\\ -0.88& 0& 0.47\end{array}$

Putting the cross products back into the Biot-savart equation we get

$\frac{{\mu }_0}{4\pi }\frac{5.4\left(2.3.{10}^{-4},0,4.4.{10}^{-4}\right)}{0.{83}^2}$

Therefore, the magnetic field at point P is

${\mathrm{dB}}_{\mathrm{x}},{\mathrm{dB}}_{\mathrm{y}},{\mathrm{dB}}_{\mathrm{z}}=\left(1.85.{10}^{-10},0,3.47.{10}^{-10}\right)\mathrm{T}$