An alpha-particle moves through a uniform magnetic field whose magnitude is 1.5 T. The field is directly parallel to the positive z-axis of the rectangular coordinate system. What is the magnitude of the magnetic force on the alpha-particle when it is moving with a velocity v = (2 i - 3 j + 1 k) x 104 m/s ?
Answer & explanation
Correct answer: option 2
F = q.(v X B)
v = (2 i - 3 j + 1 k) x 104 m/s ⇒ |v| = \(\sqrt { 2^{2} + 3^{2} + 1^{2}}\) x 104 m/s = \(\sqrt {15}\) x 104 m/s ;
B = 1.5 T ;
q = 4 e = 4 x 1.6 x 10-19 C = 6.4 x 10-19 C
F = 3.7 X 10-14 N