Charged particles are accelerated through a potential V and then pass between two charged plates, separated by a distance d and with a potential Ve between them. In the region between the plates, a uniform magnetic field B is tuned so that the particles are not deflected. Suppose now that the magnitude of B is increased by a factor of 4. By what factor must V be changed so that the particles are still not deflected? 6.AO 16 BO 1 CO 1/4 DO 1/16 EO 4 2 Solution work done by potential difference V1 = V1*q kinetic energy of particel K = (1/2)*m*v^2 work energy relation K = W (1/2)*m*v^2 = V1*q speed v = sqrt(2*V1*q/m) In between the plated electric force Fe = E*q = (V2/d)*q magnetic force Fb = q*v*B = q*sqrt(2*V1*q/m)*B for no delfection Fb = Fe q*sqrt(2*V1*q/m)*B = (V2/d)*q sqrt(2*V1*q/m)*B = (V2/d) (2*V1*q/m)*B^2 = (V2/d)^2 V1 = V2^2*m/(2*q*B^2*d^2) V1/V1\' = (B\'/B)^2 B\' = B/4 V1\' = ? V1/V1\' = 1/16 V1\' = V1*16 V1 is increased by a factor of 16 <<<----------ANSWER OPTION ( A ) .