Question

Q4. A fluorine ion (q - +4e, m 3.15 x 10-26 kg) that was accelerated through a potential difference of 48.0 kV moves in circular motion in a uniform 0.850-tesla magnetic field, see diagram. a) T rue False The diagram shows the correct direction of its motion. b) Calculate the speed of the fluorine ion. c) Calculate the radius of its cyclotron orbit.

Answer 1

a)

False. As per left hand rule.

mv2_ mv Radius of path produced by magnetic field If the velocity v is produced by an accelerating voltage V Substitution gives Magnetic field 1 /2mV ut toward observer

b)

speed of an accelerated charge is given by

v = (2qV/m)^0.5

= (2x(4x1.6x10^-19)x48000/(3.15x10^-26))^0.5

= 1396594.5 m/s

c)

r = mv/qB

= 3.15x10^-26x1396594.5 / (4x1.6x10^-19x0.85)

= 0.081m

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