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(C) The magnetic field $B$ at the center of a circular loop created by a moving charge $e$ with velocity $v$ at a distance $r$ is given by the Biot-Sa

Vedclass · Accepted Answer

$\left(\frac{\mu_{0} ev}{4 \pi B}\right)^{1 / 2}$

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(C) The magnetic field $B$ at the center of a circular loop created by a moving charge $e$ with velocity $v$ at a distance $r$ is given by the Biot-Savart law for a point charge:$B = \frac{\mu_{0}}{4 \pi} \frac{e(v 	imes r)}{r^3}$Since the velocity vector $v$ is perpendicular to the radius vector $r$ in a circular orbit,the angle $	heta = 90^{\circ}$.Thus,the magnitude is:$B = \frac{\mu_{0}}{4 \pi} \frac{ev \sin(90^{\circ})}{r^2} = \frac{\mu_{0} ev}{4 \pi r^2}$Rearranging the formula to solve for the radius $r$:$r^2 = \frac{\mu_{0} ev}{4 \pi B}$$r = \left(\frac{\mu_{0} ev}{4 \pi B}ight)^{1 / 2}$

An electron moves in a circular orbit with uniform speed $v$. It produces a magnetic field $B$ at the centre of the circle. The radius of the circle is (where $\mu_{0} =$ permeability of free space,$e =$ electronic charge):

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