A velocity selector consists of electric and magnetic fields described by the ex
ID: 1442183 • Letter: A
Question
A velocity selector consists of electric and magnetic fields described by the expressions E = E k and B = B j, with B = 11.0 mT. Find the value of E such that a 720-eV electron moving in the negative x direction is undeflected. A conductor carrying a current I = 16.3 A is directed along the positive x axis and perpendicular to a uniform magnetic field. A magnetic force per unit length of 0.124 N/m acts on the conductor in the negative y direction. Determine the magnitude of the magnetic field in the region through which the current passes. Determine the direction of the magnetic field in the region through which the current passes. +x direction -x direction +y direction -y direction +z direction -z directionExplanation / Answer
9)
energy of electron = 0.5mv^2 = 720 eV
v^2 = 2*720*1.6*10^-19 (joule)/9.1*10^-31
v = 15.91*10^6 m/s
vector > v = 15.91*10^6 (+i) >>>>>>>
for remaining undeflected > net force is zero
F-net = 0 = qE + q[v cross B] >> vector
E + [v cross B] = 0
E (k) + [v (i) cross {B} (+j)] = 0
E (k) + vB [(i) cross (+j)] = 0
E (k) + vB [k] = 0
E (k) = - vB [k] comparing magnitudes
E = vB = 15.91*10^6*11.0 *10^-3 (milli)
E = 1.75*10^5 volt/meter
10)
Force on the conducfor is F = B i l
so that B = (F/l)x 1 / i = 0.124 x 1 / 16.3 = 7.6mT
and the direction of this force is towards + Z axis, as per Fleming's left hand rule.