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(A) The electric field $E$ produced by a point charge $q$ at a distance $r$ is given by $E = \frac{1}{4 \pi \varepsilon_{0}} \frac{q}{r^{2}}$.The magn

Vedclass · Accepted Answer

$2 	imes 10^{11}$

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(A) The electric field $E$ produced by a point charge $q$ at a distance $r$ is given by $E = \frac{1}{4 \pi \varepsilon_{0}} \frac{q}{r^{2}}$.The magnetic field $B$ produced by a point charge $q$ moving with velocity $v$ at a distance $r$ is given by $B = \frac{\mu_{0}}{4 \pi} \frac{q v \sin 	heta}{r^{2}}$.For the ratio of the magnitudes, we consider the case where the velocity is perpendicular to the position vector $(\sin 	heta = 1)$.The ratio $\frac{E}{B}$ is given by:$\frac{E}{B} = \frac{\frac{1}{4 \pi \varepsilon_{0}} \frac{q}{r^{2}}}{\frac{\mu_{0}}{4 \pi} \frac{q v}{r^{2}}} = \frac{1}{\mu_{0} \varepsilon_{0} v}$.Since the speed of light $c = \frac{1}{\sqrt{\mu_{0} \varepsilon_{0}}}$, we have $c^{2} = \frac{1}{\mu_{0} \varepsilon_{0}}$.Substituting this into the ratio, we get $\frac{E}{B} = \frac{c^{2}}{v}$.Given $c = 3 	imes 10^{8} \; m/s$ and $v = 4.5 	imes 10^{5} \; m/s$:$\frac{E}{B} = \frac{(3 	imes 10^{8})^{2}}{4.5 	imes 10^{5}} = \frac{9 	imes 10^{16}}{4.5 	imes 10^{5}} = 2 	imes 10^{11} \; m/s$.

What is the ratio of the electric field to the magnetic field produced by a moving point charge if its speed is $4.5 \times 10^{5} \; m/s$?

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