A positively charged ion (charge +e; mass = 1.8 times 10^-25 kg) is accelerated

Question

A positively charged ion (charge +e; mass = 1.8 times 10^-25 kg) is accelerated through a voltage difference and undergoes the trajectory shown in the figure to right with d = 2.5 cm (please note that d is not a radius). The magnetic field is zero in the region where the ion is speeding up, and has magnitude is 3.0 T elsewhere. You may assume that the ion loses negligible amounts of energy due to dissipative forces. a. In the region with non-zero magnetic field, is the field directed into the page or out of the page? Briefly explain your reasoning. b. Is the initial voltage, V_i, positive or negative? Briefly explain your reasoning c. What is the voltage at the initial location of the ion?

Explanation / Answer

for positive charged ions,to know the direction of force,we will use Flemings Left hand rule ,which states that strecth our left hand thumb,Fore finger and middle finger are mutually perpendicular to each other

Fore finger indicated the directio of magnetic field ,middile finger gives the direction of speed of the ion and thumb gives the direction of force acting on the ion

in non zero magnetic field ,since the ion takes circular path then Field B and speed v is perpendicular to each other and force is acting towards the center of the circle ,so according to Flemings left hand rule the direction of the field is out of the page

b) to send the ion towards right as shown the initial voltage should be positive because only the repulsive force can give the push towards right side

c) radius of the orbit is r = (m*v)/(q*B)

(0.025/2) = (1.8*10^-25*v)/(1.6*10^-19*3)

v = 3.33*10^4 m/sec

Using work done through the potential difference = change in kinetic energy

q*dV = 0.5*m*v^2

1.6*10^-19(Vi) =0.5*1.8*10^-25*(3.33*10^4)^2

Vi = 623.75 V

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