Question

An elevator carrying a charge of 0.5C is moving down with a velocity of 5 x 10^{3} 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?

Moderate

Solution

According to Biot-Savart law,

$B=\frac{{\mu}_{0}}{4\pi}\frac{qv}{{r}^{2}}\mathrm{sin}\theta =\frac{{10}^{-7}\times 0.5\times 5\times {10}^{3}}{{5}^{2}}\left(\frac{3}{5}\right)=6\mu \mathrm{T}$

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Similar Questions

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 $0.5\times n\text{\hspace{0.17em}\hspace{0.17em}}\mu T$. Find n.

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