# Magnetic field due to current element

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Question

# A charge particle A of charge q = 2C has velocity V = 100 m/s. When it passes through point A and has velocity in the direction shown. The strength of instantaneous magnetic field at point B (r = 2m) due to this moving charge is  $0.5×n\text{\hspace{0.17em}\hspace{0.17em}}\mu T$. Find n.

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Solution

## $\begin{array}{l}\mathrm{d}\stackrel{\to }{\mathrm{B}}=\frac{{\mu }_{0}}{4\pi }\mathrm{I}\frac{\stackrel{\to }{\mathrm{d}\ell }×\stackrel{\to }{\mathrm{r}}}{{\mathrm{r}}^{3}}\\ \mathrm{d}\stackrel{\to }{\mathrm{B}}=\frac{{\mu }_{0}}{4\pi }\frac{\mathrm{dq}}{\mathrm{dt}}\frac{\stackrel{\to }{\mathrm{d}\ell }×\stackrel{\to }{\mathrm{r}}}{{\mathrm{r}}^{3}}\\ =\frac{{\mu }_{0}}{4\pi }\left(\mathrm{dq}\right)\left(\frac{\stackrel{\to }{\mathrm{d}\ell }}{\mathrm{dt}}\right)×\frac{\stackrel{\to }{\mathrm{r}}}{{\mathrm{r}}^{3}}\\ =\frac{{\mu }_{0}}{4\pi }\mathrm{q}\frac{\stackrel{\to }{\mathrm{u}}×\stackrel{\to }{\mathrm{r}}}{{\mathrm{r}}^{3}}=\frac{{\mu }_{0}\mathrm{qvsin}\theta }{4\pi {\mathrm{r}}^{2}}\\ =2.5×{10}^{-6}T\end{array}$

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An elevator carrying a charge of 0.5C is moving down with a velocity of 5 x 103 ms-1. The elevator is 4 m  from the bottom and 3 m horizontally away from P as shown in figure. What magnetic field (in $\mu T$) does it produce at point P?