An electric field of 1500, V/m and a magnetic | vedclass

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Question

(C) When an electron moves through both electric and magnetic fields,the net Lorentz force acting on it is given by $\vec{F} = q(\vec{E} + \vec{v} 	i

Vedclass · Accepted Answer

$3.75 	imes 10^{3} \,m/s$

Vedclass · Answer

(C) When an electron moves through both electric and magnetic fields,the net Lorentz force acting on it is given by $\vec{F} = q(\vec{E} + \vec{v} 	imes \vec{B})$.For the electron to move in a straight line at a uniform speed,the net force must be zero.This implies that the electric force must be balanced by the magnetic force: $qE = qvB$.Therefore,the speed $v$ is given by $v = \frac{E}{B}$.Given $E = 1500\, V/m$ and $B = 0.40\, Wb/m^2$,we calculate:$v = \frac{1500}{0.40} = 3750\, m/s$.This can be written in scientific notation as $v = 3.75 	imes 10^{3} \,m/s$.

An electric field of $1500\, V/m$ and a magnetic field of $0.40\, Wb/m^2$ act on a moving electron. The minimum uniform speed along a straight line the electron could have is:

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