Question

A negative charge q= -3.60 * 10-6 C is located at the origin and has velocity vS_17.50 * 104m>s2dn_1-4.90 * 104 m>s2en. At this instant what are the magnitude and direction of the magnetic field produced by this charge at the point x= 0.200 m, y= -0.300 m, z= 0?

Short answer

$\overrightarrow{B}=\left(9.75\times {10}^{-8}\mathrm{T}\right)\widehat{k}\mid$

Step-by-step solution

Concept.

Consider a negative charge$q=3.60\times {10}^{-6}\mathrm{C}$ , which is located at the origin and has a velocity of$\overrightarrow{v}=\left(7.5\times {10}^4\mathrm{m}/\mathrm{s}\right)\widehat{i}+\left(-4.90\times {10}^4\mathrm{m}/\mathrm{s}\right)\widehat{j}$ The magnetic field due to this moving electron at the nucleus, is given by,

$\overrightarrow{B}=\frac{{\mu }_o}{4\pi }\frac{q\overrightarrow{v}\times \overrightarrow{r}}{r^3}$ (1)

 Step 2: Calculation.

we need to find the magnetic field produced by this charge at the point x = 0.200 m, y= -0.300 m, and z =0. So we have $\overrightarrow{r}=\left(0.200\right)\widehat{i}+(-0.300)\widehat{j}$, thus,

$\begin{array}{c}\overrightarrow{v}\times \overrightarrow{r}=\left[\left(7.50\times {10}^4\mathrm{m}/\mathrm{s}\right)\widehat{i}+\left(-4.90\times {10}^4\mathrm{m}/\mathrm{s}\right)\widehat{j}\right]\\ \times \left[\right(0.200\mathrm{m})\widehat{i}+(-0.300\mathrm{m}\left)\widehat{j}\right]\\ =\left(-2.25\times {10}^4{\mathrm{m}}^2/\mathrm{s}\right)\widehat{k}+\left(9.80\times {10}^3{\mathrm{m}}^2/\mathrm{s}\right)\widehat{k}\\ =\left(-1.27\times {10}^4{\mathrm{m}}^2/\mathrm{s}\right)\widehat{k}\end{array}$

now we need to substitute into (1) but before we need to find r, as,

$\begin{array}{c}r=\sqrt{x^2+y^2+z^2}=\sqrt{(0.200\mathrm{m}{)}^2+(-0.300\mathrm{m}{)}^2}=0.3606\mathrm{m}\\ \overrightarrow{B}=\frac{\left(1.00\times {10}^{-7}\mathrm{T}\cdot \mathrm{m}/\mathrm{A}\right)\left(-3.60\times {10}^{-6}\mathrm{C}\right)\left(-1.27\times {10}^4{\mathrm{m}}^2/\mathrm{s}\right)}{(0.3606\mathrm{m}{)}^3}\widehat{k}\\ =\left(9.75\times {10}^{-8}\mathrm{T}\right)\widehat{k}\\ \overrightarrow{B}=\left(9.75\times {10}^{-8}\mathrm{T}\right)\widehat{k}\end{array}$