Answer Happy

p8,6
for second question
m=3.2×10-²⁶ kg
q=1.6×10‐¹⁹ C
v=459.43 v
B=1.58T
R?
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18) A particle with charge q=2 HC enters at point A into region with a uniform normal magnetic field B-0.04T,as shown in the diagram. The momentum (in kg.m/s) of the particle at point B is Point D xxxx xxxx xxxxxx xxxxx (A)1.6x10-8 (B) 3.2x10-8 (C) 6.4x10-8 (D) 1.28x10-7 (E) 0.8x10-8 T. Hasan Abdulla Salman A singly charged positive ion has a mass of 3.20 10-26 kg and charge of 1.6 x 10:19 C. After being accelerated from rest through a potential difference of 459.43 V, the ion enters a magnetic field of 1.50 T along a direction perpendicular to the direction of the field, as shown in the figure. Calculate the radius of the path of the ion in the field. R=0.2 m
Q5: An ion is moving with velocity vo enters a uniform magnetic field B directed into the page. If the ion is deflected as shown in the figure, then the magnitude of the ratio of its charge to mass is, (A) 10v/B ((B)20v/B (C) vo/B (D2v/B X X X X X X X B X X (into) X X X X X X -10.0 cm- (E) 5V/B