An electron (e^-) is moving parallel to a lon | vedclass

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(B) The magnetic field $(B)$ produced by a long straight wire at a distance $(r)$ is given by $B = \frac{\mu_0 I}{2 \pi r}$.Substituting the values: $

Vedclass · Accepted Answer

$0.8 	imes 10^{-18} \ N$

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(B) The magnetic field $(B)$ produced by a long straight wire at a distance $(r)$ is given by $B = \frac{\mu_0 I}{2 \pi r}$.Substituting the values: $B = \frac{2 	imes 10^{-7} 	imes 10}{0.04} = 0.5 	imes 10^{-4} \ T$.The magnetic force $(F_m)$ on a moving charge is $F_m = qvB \sin 	heta$.Here,the velocity is parallel to the wire,so the angle $(	heta)$ between velocity and magnetic field is $90^{\circ}$.$F_m = (1.6 	imes 10^{-19} \ C) 	imes (10^5 \ m/s) 	imes (0.5 	imes 10^{-4} \ T) 	imes \sin 90^{\circ}$.$F_m = 1.6 	imes 10^{-19} 	imes 10^5 	imes 0.5 	imes 10^{-4} = 0.8 	imes 10^{-18} \ N$.

An electron $(e^-)$ is moving parallel to a long current-carrying wire as shown in the figure. Calculate the magnetic force acting on the electron. (Given: $I = 10 \ A$,$v = 10^5 \ m/s$,$r = 4 \ cm = 0.04 \ m$)

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